287. Find the Duplicate Number

https://leetcode.com/problems/find-the-duplicate-number/#/description

http://www.cnblogs.com/EdwardLiu/p/5078013.html

range search : The reason why we did not use index as "search space" for this problem is the matrix is sorted in two directions, we can not find a linear way to map the number and its index.------------(Unsorted Array)

时间: 2024-10-28 15:43:33

287. Find the Duplicate Number的相关文章

<LeetCode OJ> 287. Find the Duplicate Number

287. Find the Duplicate Number My Submissions Question Total Accepted: 18097 Total Submissions: 48596 Difficulty: Hard Given an array nums containing n + 1 integers where each integer is between 1 and n (inclusive), prove that at least one duplicate

LeetCode 287. Find the Duplicate Number (python 判断环,时间复杂度O(n))

LeetCode 287. Find the Duplicate Number 暴力解法 时间 O(nlog(n)),空间O(n),按题目中Note"只用O(1)的空间",照理是过不了的,但是可能判题并没有卡空间复杂度,所以也能AC. class Solution: # 基本思路为,将第一次出现的数字 def findDuplicate(self, nums: List[int]) -> int: s = set() for i in nums: a = i in s if a

leetcode 217. Contains Duplicate 287. Find the Duplicate Number

217. Contains Duplicate 后面3个题都是限制在1-n的 class Solution { public: bool containsDuplicate(vector<int>& nums) { int length = nums.size(); if(length <= 0) return false; sort(nums.begin(),nums.end()); for(int i = 1;i < length;i++){ if(nums[i] ==

LeetCode 287. Find the Duplicate Number

Find the Duplicate Number | LeetCode OJhttps://leetcode.com/problems/find-the-duplicate-number/ 这个题目属于编码比较简单但解法分析过程比较复杂. 首先,把1~n放入0~n个元素,必定有两个或以上元素重复.数字里没有0,所以从下标0出发不会再回到最初的元素0. 假设当前的下标为x,下一步的下标为f(x),若f(x) = A[x], 也即每次跳到一个元素,则下一步移动到当前元素值对应的元素下标.我们来证明

LeetCode 287. Find the Duplicate Number (找到重复的数字)

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 th

[LeetCode] 287. Find the Duplicate Number 寻找重复数

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: [1,3,4,2,2

[LeetCode] 287. Find the Duplicate Number(Floyd判圈算法)

传送门 Description 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 mu

287. Find the Duplicate Number *HARD*

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 th

leetcode 287 Find the Duplicate Number寻找重复数

这道题用STL容器就很好写了,可以用set也可以用map, 用unordered_map的C++代码如下: 1 class Solution { 2 public: 3 int findDuplicate(vector<int>& nums) { 4 unordered_map<int, int> m; 5 int res; 6 for(int i=0;i<nums.size();i++){ 7 if(m.count(nums[i])){ 8 res=nums[i];