Leetcode 287. Find the Duplicate Number Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Note: You must not modify...
https://leetcode.com/problems/find-the-duplicate-number/description/ 287. Find the Duplicate Number Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate ...
287. Find the Duplicate Number Medium Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate number must exist. Assume that there is only one duplicate number, find the duplicate one. Example 1: Input: [1,3,4...
classSolution {public:intfindDuplicate(vector<int>&nums) {intleft =1, right =nums.size();while(left<right){intmid = left+(right-left)/2,cnt=0;for(intnum:nums){if(num<=mid)++cnt; }if(cnt<=mid)left=mid+1;elseright=mid; }returnright; } }; Python3 #Python3classSolution:deffindD...
题目原址 https://leetcode.com/problems/find-the-duplicate-number/description/ 题目描述 Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one ...LeetCode 287. Find the Duplicate Number Given an array nums containing n +...
Can you solve this real interview question? Find the Duplicate Number - Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive. There is only one repeated number in nums, return this repeated number
func findDuplicate(nums []int) int { // 二分区间左边界,初始化为 1 l := 1 // 二分区间右边界,初始化为 n r := len(nums) - 1 // 当前区间不为空时,继续二分 for l <= r { // 计算区间中点 mid mid := (l + r) >> 1 // 统计 nums 中小于等于 mid 的数字个数 count := 0...
Leetcode每日一题:287.find-the-duplicate-number(寻找重复数),思路:一开始并没有什么头绪,直接排序加遍历以O(nlgn)的复杂度水过去了,后来看评论才知道有Floyd判圈算法这么秒的方法,简称龟兔赛跑;具体算法讲解可参考文章:算法-floyd判环(圈)算法,讲得很棒,便于理
// https://leetcode.cn/problems/find-the-duplicate-number/solutions/7038/er-fen-fa-si-lu-ji-dai-ma-python-by-liweiwei1419/classSolution{publicintfindDuplicate(int[]nums){intlen=nums.length;// n + 1 = len, n = len - 1// 在 [1..n] 查找 nums 中重复的元素intleft=1;intright=len...
代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{publicintfindDuplicate(int[]nums){Set<Integer>seen=newHashSet<>();for(int n:nums){if(seen.contains(n)){returnn;}seen.add(n);}return-1;}} 复杂度分析