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【1010cc时时彩经典版】面试中常见算法难题详解

发布时间:2019-09-30 15:34编辑:1010cc时时彩客户端浏览(184)

    数组去重

    给定某冬日数组,供给删除数组中的重复数字况兼重临新的无重复数组。

    JavaScript

    // ES6 Implementation var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8]; Array.from(new Set(array)); // [1, 2, 3, 5, 9, 8] // ES5 Implementation var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8]; uniqueArray(array); // [1, 2, 3, 5, 9, 8] function uniqueArray(array) { var hashmap = {}; var unique = []; for(var i = 0; i < array.length; i ) { // If key returns null (unique), it is evaluated as false. if(!hashmap.hasOwnProperty([array[i]])) { hashmap[array[i]] = 1; unique.push(array[i]); } } return unique; }

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    // ES6 Implementation
    var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];
     
    Array.from(new Set(array)); // [1, 2, 3, 5, 9, 8]
     
     
    // ES5 Implementation
    var array = [1, 2, 3, 5, 1, 5, 9, 1, 2, 8];
     
    uniqueArray(array); // [1, 2, 3, 5, 9, 8]
     
    function uniqueArray(array) {
      var hashmap = {};
      var unique = [];
      for(var i = 0; i < array.length; i ) {
        // If key returns null (unique), it is evaluated as false.
        if(!hashmap.hasOwnProperty([array[i]])) {
          hashmap[array[i]] = 1;
          unique.push(array[i]);
        }
      }
      return unique;
    }

    7# Leetcode 69. Sqrt(x)

    Implement int sqrt(int x).
    Compute and return the square root of x.

    public class Solution {
        public int mySqrt(int x) {
            long start = 1;
            long end = x;
            while (start   1 < end) {
                long mid = start   (end - start) / 2;
                if(mid * mid <= x) {
                    start = mid;
                }
                else {
                    end = mid;
                }
            }
            if(end * end <= x) {
                return (int)end;
            }
            return (int)start;
        }
    }
    

    做的片段题的解题思路

    function insertData(name,lastName,phone,address){
    code here;
    }

    //Example 1
    insertData({name:'Mike', lastName:'Rogers', phone:'555-555-5555',address:'the address', status:'married'});

    JavaScript 面试中常见算法难题详解

    2017/02/20 · JavaScript · 1 评论 · 算法

    原著出处: 王下邀月熊_Chevalier   

    JavaScript 面试中常见算法难题详解 翻译自 Interview Algorithm Questions in Javascript() {…} 从属于小编的 Web 前端入门与工程举办。下文提到的广大难点从算法角度并不应当要么困难,不过用 JavaScript 内置的 API 来完毕大概必要一番勘察的。

    40# LintCode: Maximum Number in Mountain Sequence

    Given a mountain sequence of n integers which increase firstly and then decrease, find the mountain top.

    Example
    Given nums = [1, 2, 4, 8, 6, 3] return 8
    Given nums = [10, 9, 8, 7], return 10

    public int mountainSequence(int[] nums) {
        if (nums.length == 0 || nums == null) {
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        while (start   1 < end) {
            int mid = start   (end - start) / 2;
            if (nums[mid] > nums[mid   1]) {
                end = mid;
            } else {
                start = mid;
            }
        }
        return Math.max(nums[start], nums[end]);
    }
    

    Evaluate Reverse Polish Notation

    计量逆波兰共和国表明式(又叫后缀说明式)的值

    '' 2 '','' 1 '','' '', ''3'', ''* '' -->(2 1)*3-->9

    用货仓碰着运算符则把前边多个拿出来运算

    public class Main {
        public static void main(String[] args) {
           String []tokens={"2", "1", " ", "3", "*"};
            System.out.print(evalRPN(tokens));
        }
    
        public static int evalRPN(String [] tokens){
            Stack<String> s = new Stack<>();
            if(tokens.length==1){
                return Integer.parseInt(tokens[0]);
            }
            for(String token:tokens){
                if(!isOperator(token)){
                    s.push(token);
                }else {
                    int y=Integer.parseInt(s.pop());
                    int x=Integer.parseInt(s.pop());
                    switch (token.charAt(0)){
                        case ' ':x =y;break;
                        case '-':x-=y;break;
                        case '*':x*=y;break;
                        case '/':x/=y;break;
                    }
                    s.push(String.valueOf(x));
                }
            }
            return Integer.parseInt(s.peek());
        }
    
        private static boolean isOperator(final String op){
            return op.length() == 1 && OPS.indexOf(op)!=-1;
        }
    
        private static String OPS = new String(" -*/");
    }
    

    ...

    复制代码 代码如下:

    会问字符串

    认清有个别字符串是或不是为回文字符串,例如racecarrace car都以回文字符串:

    JavaScript

    isPalindrome("racecar"); // true isPalindrome("race Car"); // true function isPalindrome(word) { // Replace all non-letter chars with "" and change to lowercase var lettersOnly = word.toLowerCase().replace(/s/g, ""); // Compare the string with the reversed version of the string return lettersOnly === lettersOnly.split("").reverse().join(""); }

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    isPalindrome("racecar"); // true
    isPalindrome("race Car"); // true
     
    function isPalindrome(word) {
      // Replace all non-letter chars with "" and change to lowercase
      var lettersOnly = word.toLowerCase().replace(/s/g, "");
     
      // Compare the string with the reversed version of the string
      return lettersOnly === lettersOnly.split("").reverse().join("");
    }

    39# LintCode: Last Position of Target

    Find the last position of a target number in a sorted array. Return -1 if target does not exist.
    Example
    Given [1, 2, 2, 4, 5, 5].
    For target = 2, return 2.
    For target = 5, return 5.
    For target = 6, return -1.

    public int lastPosition(int[] nums, int target) {
        // Write your code here
        if (nums.length == 0 || nums == null) {
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        while (start   1 < end) {
            int mid = start   (end - start) / 2;
            if (nums[mid]==target) {
                start = mid;
            } else if (nums[mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (nums[end] == target) {
            return end;
        }
        if (nums[start] == target) {
            return start;
        }
        return -1;
    }
    

    Contains Duplicate II

    Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k.

    维护一个HashMap,key为整数,value为下标,将数组中的元素不断添加到这个Hashmap中,遇到重复时,计算下标距离;
    用Integer.MAX_VALUE 设置为比较的初始值;
    学会用HashMap是非常关键的。

    public class LeetCode {
        public static void main(String[] args) {
            int[] nums = {1, 3, 5, 1,6};
            int k=3;
            System.out.print(containsNearbyDuplicate(nums,k));
        }
    
        public static boolean containsNearbyDuplicate(int[] nums, int k) {
            final Map<Integer,Integer> map = new HashMap<>();
            int min=Integer.MAX_VALUE;
    
            for(int i=0;i<nums.length;i  ){
                if(map.containsKey(nums[i])){
                    final int preIndex=map.get(nums[i]);
                    final int gap = i-preIndex;
                    min = Math.min(min,gap);
                }
                map.put(nums[i],i);
            }
            return min<=k;
        }
    }
    

    复制代码 代码如下:

    9,增加本地对象
    即使如此片段JavaScript的元首不推荐那个手艺,可是它早就被部分框架使用了。它能够让您针对JavaScript API来成立一些帮助性的点子。

    数字

    23# Leetcode 436. Find Right Interval

    Given a set of intervals, for each of the interval i, check if there exists an interval j whose start point is bigger than or equal to the end point of the interval i, which can be called that j is on the "right" of i.

    For any interval i, you need to store the minimum interval j's index, which means that the interval j has the minimum start point to build the "right" relationship for interval i. If the interval j doesn't exist, store -1 for the interval i. Finally, you need output the stored value of each interval as an array.

    Note:
    You may assume the interval's end point is always bigger than its start point.
    You may assume none of these intervals have the same start point.
    Example 1:
    Input: [ [1,2] ]
    Output: [-1]
    Explanation: There is only one interval in the collection, so it outputs -1.

    Example 2:
    Input: [ [3,4], [2,3], [1,2] ]
    Output: [-1, 0, 1]
    Explanation: There is no satisfied "right" interval for [3,4].
    For [2,3], the interval [3,4] has minimum-"right" start point;
    For [1,2], the interval [2,3] has minimum-"right" start point.

    Example 3:
    Input: [ [1,4], [2,3], [3,4] ]
    Output: [-1, 2, -1]
    【1010cc时时彩经典版】面试中常见算法难题详解,JavaScript初学者须求精晓13个小技术。Explanation: There is no satisfied "right" interval for [1,4] and [3,4].
    For [2,3], the interval [3,4] has minimum-"right" start point.

    Increasing Triplet Subsequence

    Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

    Formally the function should:
    Return true if there exists i, j, k
    such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
    Your algorithm should run in O(n) time complexity and O(1) space complexity.

    Examples:
    Given [1, 2, 3, 4, 5],
    return true.

    Given [5, 4, 3, 2, 1],
    return false.
    用整数最大值去比较,x1记录第一个数,x2记录第二大的数,当出现第三大的数,则return true。

    public class Solution {
        public boolean increasingTriplet(int[] nums) {
            int x1=Integer.MAX_VALUE;
            int x2=Integer.MAX_VALUE;
    
            for(int x: nums){
                if(x<=x1) x1=x;
                else if(x<=x2) x2=x;
                else return true;
            }
            return false;
        }
    }
    

    function add_nums(){
    return arguments.length;
    }
    add_nums(23,11,32,56,89,89,89,44,6); //this return the number 9

    【1010cc时时彩经典版】面试中常见算法难题详解,JavaScript初学者须求精晓13个小技术。复制代码 代码如下:

    数组中元素最大差值总计

    给定某无序数组,求取任意五个要素之间的最大差值,注意,这里须要差值总结中相当的小的成分下标必须低于十分的大体素的下标。举例[7, 8, 4, 9, 9, 15, 3, 1, 10]这些数组的总结值是 11( 15 – 4 ) 并非 14(15 – 1),因为 15 的下标小于 1。

    JavaScript

    var array = [7, 8, 4, 9, 9, 15, 3, 1, 10]; // [7, 8, 4, 9, 9, 15, 3, 1, 10] would return `11` based on the difference between `4` and `15` // Notice: It is not `14` from the difference between `15` and `1` because 15 comes before 1. findLargestDifference(array); function findLargestDifference(array) { // 借使数组只有贰个要素,则一直回到 -1 if (array.length <= 1) return -1; // current_min 指向当前的小不点儿值 var current_min = array[0]; var current_max_difference = 0; // 遍历整个数组以求取当前最大差值,借使发现有个别最大差值,则将新的值覆盖 current_max_difference // 同一时候也会追踪当前数组中的最小值,进而确定保障 `largest value in future` - `smallest value before it` for (var i = 1; i < array.length; i ) { if (array[i] > current_min && (array[i] - current_min > current_max_difference)) { current_max_difference = array[i] - current_min; } else if (array[i] <= current_min) { current_min = array[i]; } } // If negative or 0, there is no largest difference if (current_max_difference <= 0) return -1; return current_max_difference; }

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    var array = [7, 8, 4, 9, 9, 15, 3, 1, 10];
    // [7, 8, 4, 9, 9, 15, 3, 1, 10] would return `11` based on the difference between `4` and `15`
    // Notice: It is not `14` from the difference between `15` and `1` because 15 comes before 1.
     
    findLargestDifference(array);
     
    function findLargestDifference(array) {
     
      // 如果数组仅有一个元素,则直接返回 -1
     
      if (array.length <= 1) return -1;
     
      // current_min 指向当前的最小值
     
      var current_min = array[0];
      var current_max_difference = 0;
      
      // 遍历整个数组以求取当前最大差值,如果发现某个最大差值,则将新的值覆盖 current_max_difference
      // 同时也会追踪当前数组中的最小值,从而保证 `largest value in future` - `smallest value before it`
     
      for (var i = 1; i < array.length; i ) {
        if (array[i] > current_min && (array[i] - current_min > current_max_difference)) {
          current_max_difference = array[i] - current_min;
        } else if (array[i] <= current_min) {
          current_min = array[i];
        }
      }
     
      // If negative or 0, there is no largest difference
      if (current_max_difference <= 0) return -1;
     
      return current_max_difference;
    }

    36# Leetcode 302.Smallest Rectangle Enclosing Black Pixels

    Add Two Numbers

    You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

    You may assume the two numbers do not contain any leading zero, except the number 0 itself.

    Input: (2 -> 4 -> 3) (5 -> 6 -> 4)
    Output: 7 -> 0 -> 8
    主要学习怎么创建链表,怎么定义链表

    public class LeetCode {
        public static void main(String[] args) {
            int[] inputl1=new int[]{2,4,3};
            int[] inputl2=new int[]{5,6,4};
            ListNode l1=buildListNode(inputl1);
            ListNode l2=buildListNode(inputl2);
            ListNode listNode =addTwoNumbers(l1,l2);
            while(listNode!=null){
                System.out.println("val " listNode.val);
                listNode=listNode.next;
            }
        }
        //定义链表
       public static class ListNode{
            int val;
            ListNode next;
            ListNode(int val){
                this.val=val;
                this.next=null;
            }
        }
        //创建链表
        private static ListNode buildListNode(int[] input){
            ListNode first = null,last = null,newNode;
            int num;
            if(input.length>0){
                for(int i=0;i<input.length;i  ){
                    newNode=new ListNode(input[i]);
                    newNode.next=null;
                    if(first==null){
                        first=newNode;
                        last=newNode;
                    }
                    else{
                        last.next=newNode;
                        last=newNode;
                    }
                }
            }
            return first;
        }
    
        /**
         * Definition for singly-linked list.
         * public class ListNode {
         *     int val;
         *     ListNode next;
         *     ListNode(int x) { val = x; }
         * }
         */
            public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
               ListNode dummy =new ListNode(-1);
               int carry = 0;
               ListNode prev = dummy;
               for(ListNode pa=l1 ,  pb=l2 ; pa!=null || pb!=null ;
                pa=pa==null?null : pa.next,
                pb=pb==null ? null : pb.next,
                prev=prev.next){
                   final int ai=pa==null?0:pa.val;
                   final int bi=pb==null?0:pb.val;
                   final int value=(ai bi carry);
                   carry=(ai bi carry)/10;
                   prev.next=new ListNode (value);
                }
    
                if(carry>0)
                    prev.next=new ListNode (carry);
                return dummy.next;
        }
    }
    

    9,扩大本地对象
    尽管部分JavaScript的元首不引入那一个技术,可是它曾经被一些框架使用了。它能够令你针对JavaScript API来创立一些协理性的方法。

    //Example 2
    var myData = { name:'Mike',
    lastName:'Rogers',
    phone:'555-555-5555',
    address:'the address',
    status:'married'
    };
    insertData(myData);

    JavaScript Specification

    11# Leetcode 350. Intersection of Two Arrays II

    Given two arrays, write a function to compute their intersection.

    Example:
    Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

    Note:
    Each element in the result should appear as many times as it shows in both arrays.
    The result can be in any order.
    Follow up:
    What if the given array is already sorted? How would you optimize your algorithm?
    What if nums1's size is small compared to nums2's size? Which algorithm is better?
    What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

    public class Solution {
        public int[] intersect(int[] nums1, int[] nums2) {
            Arrays.sort(nums1);
            Arrays.sort(nums2);
            int index1 = 0;
            int index2 = 0;
            List<Integer> list = new ArrayList<>();
            while(index1 < nums1.length && index2 < nums2.length) {
                if (nums1[index1] == nums2[index2]) {
                    list.add(nums1[index1]);
                    index1  ;
                    index2  ;
                } else if (nums1[index1] < nums2[index2]) {
                    index1  ;
                } else if (nums1[index1] > nums2[index2]) {
                    index2  ;
                }
            }
            int[] result = new int[list.size()];
            int index = 0;
            for (int element: list) {
                result[index  ] = element;
            }
            return result;
        }
    }
    

    Product of Array Except Self

    除自身之外的数组之积
    Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
    Solve it without division and in O(n).
    For example, given [1,2,3,4], return [24,12,8,6].

    解题思路:拆分法
    [A_234 , A_134 , A_124 , A_123 ]=
    [1 , A_1 , A_12 , A_123 ]*
    [A_234 , A_34 , A_4 , 1 ]

    /**
     * Created by Administrator on 2017/5/8.
     */
    public class LeetCode {
    
        public static void main(String[] args) {
            // int [] nums={5, 7, 1, 8,3, 10};  //测试
            int[] nums = {1, 3, 5, 6};
            int k = 5;
            int [] res = productExceptSelf(nums);
            for (int i=0;i<res.length;i  ) {
                System.out.print(res[i] " ");
            }
        }
    
        public static int[] productExceptSelf(int[] nums) {
            final int [] result = new int [nums.length];
            final int [] right = new int [nums.length];
            final int [] left = new int [nums.length];
            left[0]=1;
            for(int i=1;i<nums.length;i  ){
                left[i]=left[i-1]*nums[i-1];
            }
            right[nums.length-1]=1;
            for(int i=nums.length-2;i>=0;i--){
                right[i]=right[i 1]*nums[i 1];
            }
    
            for (int i=0;i<nums.length;i  ){
                result[i]=right[i]*left[i];
            }
            return  result;
        }
    }
    

    var mynumber = 234;
    typeof mynumber; //Number
    mynumber = '';
    typeof mynumber; //String

    function insertData(name,lastName,phone,address){
    code here;
    }

    == 与 === 的分别是什么

    === 也正是所谓的严加相比较,关键的分别在于=== 会相同的时间比较类型与值,实际不是仅相比较值。

    JavaScript

    // Example of comparators 0 == false; // true 0 === false; // false 2 == '2'; // true 2 === '2'; // false

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    // Example of comparators
    0 == false; // true
    0 === false; // false
     
    2 == '2'; // true
    2 === '2'; // false

    16# Leetcode 34. Search for a Range

    Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.

    Your algorithm's runtime complexity must be in the order of O(log n).

    If the target is not found in the array, return [-1, -1].

    For example,
    Given [5, 7, 7, 8, 8, 10] and target value 8,
    return [3, 4].

    public class Solution {
        public int[] searchRange(int[] nums, int target) {
           if (nums.length == 0) {
                return new int[]{-1,-1};
            }
            int[] bound = new int[2];
            // search for left bound
            int start = 0;
            int end = nums.length - 1;
            while (start   1 < end) {
                int mid = start   (end - start) / 2;
                if (nums[mid] == target) {
                    end = mid;
                } else if (nums[mid] < target) {
                    start = mid;
                } else {
                    end = mid;
                }
            }
            if (nums[start] == target) {
                bound[0] = start;
            } else if (nums[end] == target) {
                bound[0] = end;
            } else {
                bound[0] = bound[1] = -1;
                return bound;
            }
            //search for right bound
            start = 0;
            end = nums.length - 1;
            while (start   1 < end) {
                int mid = start   (end - start) / 2;
                if (nums[mid] == target) {
                    start = mid;
                } else if (nums[mid] < target) {
                    start = mid;
                } else {
                    end = mid;
                }
            }
            if (nums[end] == target) {
                bound[1] = end;
            } else if (nums[start] == target) {
                bound[1] = start;
            } else {
                bound[0] = bound[1] = -1;
                return bound;
            }
            return bound;
        }
    }
    

    复制代码 代码如下:

    function add_nums(num1, num2){
    return num1 num2;
    }
    add_nums.length // 2 is the amount of parameters expected by the function add_nums

    数组交集

    给定四个数组,须求求出四个数组的交集,注意,交聚焦的因素应该是天下无敌的。

    JavaScript

    var firstArray = [2, 2, 4, 1]; var secondArray = [1, 2, 0, 2]; intersection(firstArray, secondArray); // [2, 1] function intersection(firstArray, secondArray) { // The logic here is to create a hashmap with the elements of the firstArray as the keys. // After that, you can use the hashmap's O(1) look up time to check if the element exists in the hash // If it does exist, add that element to the new array. var hashmap = {}; var intersectionArray = []; firstArray.forEach(function(element) { hashmap[element] = 1; }); // Since we only want to push unique elements in our case... we can implement a counter to keep track of what we already added secondArray.forEach(function(element) { if (hashmap[element] === 1) { intersectionArray.push(element); hashmap[element] ; } }); return intersectionArray; // Time complexity O(n), Space complexity O(n) }

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    var firstArray = [2, 2, 4, 1];
    var secondArray = [1, 2, 0, 2];
     
    intersection(firstArray, secondArray); // [2, 1]
     
    function intersection(firstArray, secondArray) {
      // The logic here is to create a hashmap with the elements of the firstArray as the keys.
      // After that, you can use the hashmap's O(1) look up time to check if the element exists in the hash
      // If it does exist, add that element to the new array.
     
      var hashmap = {};
      var intersectionArray = [];
     
      firstArray.forEach(function(element) {
        hashmap[element] = 1;
      });
     
      // Since we only want to push unique elements in our case... we can implement a counter to keep track of what we already added
      secondArray.forEach(function(element) {
        if (hashmap[element] === 1) {
          intersectionArray.push(element);
          hashmap[element] ;
        }
      });
     
      return intersectionArray;
     
      // Time complexity O(n), Space complexity O(n)
    }

    37# Leetcode 174. Dungeon Game

    The demons had captured the princess (P) and imprisoned her in the bottom-right corner of a dungeon. The dungeon consists of M x N rooms laid out in a 2D grid. Our valiant knight (K) was initially positioned in the top-left room and must fight his way through the dungeon to rescue the princess.

    The knight has an initial health point represented by a positive integer. If at any point his health point drops to 0 or below, he dies immediately.

    Some of the rooms are guarded by demons, so the knight loses health (negative integers) upon entering these rooms; other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers).

    In order to reach the princess as quickly as possible, the knight decides to move only rightward or downward in each step.

    Write a function to determine the knight's minimum initial health so that he is able to rescue the princess.

    For example, given the dungeon below, the initial health of the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN.

    -2(K) -3 3
    -5 -10 1
    10 30 -5(P)

    Notes:

    The knight's health has no upper bound.
    Any room can contain threats or power-ups, even the first room the knight enters and the bottom-right room where the princess is imprisoned.

    2,调解三个数组的尺寸
    Length属性不是只读的,所以您能够设置Length属性的值。并且,你能够运用它增大或降低数组的长度。譬如:

    var myString = '23255';
    typeof myString; //String
    myString = !!myString;
    typeof myString //Boolean

    解释下 Prototypal Inheritance 与 Classical Inheritance 的区别

    在类承袭中,类是不可变的,分裂的语言中对此多再三再四的支撑也不一样,有些语言中还补助接口、final、abstract 的概念。而原型承接则越是灵活,原型本身是能够可变的,并且对象只怕承袭自八个原型。

    20# Leetcode 378. Kth Smallest Element in a Sorted Matrix

    复制代码 代码如下:

    现行反革命,要利用这么些函数拾壹分的简练;大家可以用二种艺术来发送数据:

    解释下 null 与 undefined 的区别

    JavaScript 中,null 是一个足以被分配的值,设置为 null 的变量意味着其无值。而 undefined 则表示着某些变量即使声称了可是从未开展过其余赋值。

    15# Leetcode 81. Search in Rotated Sorted Array II

    Follow up for "Search in Rotated Sorted Array":
    What if duplicates are allowed?

    Would this affect the run-time complexity? How and why?
    Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

    Write a function to determine if a given target is in the array.

    The array may contain duplicates.

    public class Solution {
        // 这个问题在面试中不会让实现完整程序
        // 只需要举出能够最坏情况的数据是 [1,1,1,1... 1] 里有一个0即可。
        // 在这种情况下是无法使用二分法的,复杂度是O(n)
        // 因此写个for循环最坏也是O(n),那就写个for循环就好了
        //  如果你觉得,不是每个情况都是最坏情况,你想用二分法解决不是最坏情况的情况,那你就写一个二分吧。
        //  反正面试考的不是你在这个题上会不会用二分法。这个题的考点是你想不想得到最坏情况。
        public boolean search(int[] nums, int target) {
            for (int i = 0; i < nums.length; i   ) {
                if (nums[i] == target) {
                    return true;
                }
            }
            return false;
        }
    }
    

    复制代码 代码如下:

    复制代码 代码如下:

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