Find the Duplicate Number
来源:互联网 发布:vb 修改网页源代码 编辑:程序博客网 时间:2024/06/05 20:31
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 only).
You must use only constant, O(1) extra space.
Your runtime complexity should be less than O(n2).
There is only one duplicate number in the array, but it could be repeated more than once.
方法:参考https://leetcode.com/problems/linked-list-cycle-ii/
寻找进入还的第一个节点。
class Solution {public: int findDuplicate(vector<int>& nums) { if(nums.size()<1) return -1; int slow = 0, fast = 0; while(slow==0 || slow != fast){ slow = nums[slow]; fast = nums[nums[fast]]; } int cur = 0; while(cur != slow){ cur = nums[cur]; slow = nums[slow]; } return slow; }};
0 0
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find The Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- Find the Duplicate Number
- 拷贝头文件shell
- Win10 64bit配置Selenium2环境
- Android屏幕、键盘背光Framework和Linux led_classdev
- Spring+SpringMVC+Mybatis整合系列(二)Eclipse新建Maven web项目
- 谷歌浏览器慎用有道词典插件(<audio></audio>)
- Find the Duplicate Number
- 最全的Java笔试题库之选择题篇-总共234道【61~120】
- 1 NXP的BLE协议栈软件架构与应用层代码分析
- 4866: [Ynoi2017]由乃的商场之旅
- [leetcode]: 122. Best Time to Buy and Sell Stock II
- POJ3421_X-factor Chains_素因子分解&&排列组合
- 云服务器lnmp环境一键安装出现错误
- SpringMVC注解配置Swagger2步骤及相关知识点
- centos版本阿里云服务器安装naginx以及相关配置