LeetCode Top Interview Questions 217. Contains Duplicate (Java版; Easy) 题目描述 Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least
LeetCode Top 100 Liked Questions 53. Maximum Subarray (Java版; Easy) 题目描述 Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. Example: Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Explanation:...
https://leetcode.cn/problems/minimum-time-to-make-array-sum-at-most-x 这道题代码很好写,但思路非常难想到。重复操作相同下标一定是血亏的,而根据排序不等式,如果选了一些下标j,则对应nums2元素较大的j应该尽量晚操作。因此正解是转化为二维DP,状态定义为前i个数(注意是按nums2从小到大排好的前i个)中...
[LeetCode] Top 100 Liked Questions All LeetCode Questions List 题目汇总
1 <= nums.length <= 105 -104<= nums[i] <= 104 kis in the range[1, the number of unique elements in the array]. It isguaranteedthat the answer isunique. Follow up:Your algorithm's time complexity must be better thanO(n log n), where n is the array's size....
LeetCode 347题目可以使用哪些排序算法来解决? 题目: 给定一个非空的整数数组,返回其中出现频率前 K 高的元素。 Given a non-empty array of integers, return the K most frequent elements. 示例1: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 输入: nums = [1,1,1,2,2,3], k = 2 输出...
448. Find All Numbers Disappeared in an Array Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space...
LeetCode 指南语言: Java 说明: 每道题在代码头部都添加了我的解题思路和批注,Eg: /*** * 287. Find the Duplicate Number * 题意:n+1个数属于[1~n],找出重复的那个数 * 难度:Medium * 分类:Array, Two Pointers, Binary Search * 思路:如果nums[i]不在对应位置,则和对应位置交换。如果对应位置上...
200 LeetCode Top Interview Questions: Optimal Solutions with Detailed Explanations for Easy/Medium/Hard Levels to Ace Coding Interviews downdemo.github.io/LeetCode-Solutions-in-Cpp17/ Topics data-structures-and-algorithms Resources Readme License MIT license Activity Stars 63 stars Watchers ...
11. Missing numberhttps://leetcode.com/problems/missing-number Given an arraynumscontainingndistinct numbers in the range[0, n], returnthe only number in the range that is missing from the array. sum(0:n) - sum(nums[0]:nums[n-1]) ...