LeetCode 287. Find the Duplicate Number (找到重复的数字) Given an arraynumscontainingn+ 1 integers where each integer is between 1 andn(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 n...
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 the array (assume the array is read o...
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...
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
Assume that there is only one duplicate number, find the duplicate one. 即一个包含n+1个整型的数组,数组元素的大小范围是[1,n],说明至少有一个重复的元素。假设只有一个重复元素,找出这个元素。 Example 1: Input: [1,3,4,2,2] Output: 2 Example 2: Input: [3,1,3,4,2] Output: 3 ...
class Solution: def findDuplicate(self, nums: List[int]) -> int: # 二分区间左边界,初始化为 1 l: int = 1 # 二分区间右边界,初始化为 n r: int = len(nums) - 1 # 当前区间不为空时,继续二分 while l <= r: # 计算区间中点 mid mid: int = (l + r) >> 1 # 统计 nums 中小...
算法: public int findDuplicate(int[] nums) { int n = nums.length-1,l=1,r=n; while(l<r){ int m = (l+r)/2,count=0,t=0; for(int i=0;i<nums.length;i++){ if(nums[i]>m) count++; if(nums[i]==m){ t++; }
public int findDuplicate(int[] nums) { int slow = 0; int fast = 0; // 找到快慢指针相遇的地方 do{ slow = nums[slow]; fast = nums[nums[fast]]; } while(slow != fast); int find = 0; // 用一个新指针从头开始,直到和慢指针相遇 ...
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
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: ...