【latched】Single Number II
来源:互联网 发布:百度助手 网络不可用 编辑:程序博客网 时间:2024/06/05 23:56
Problem
Code
int singleNumber(int* nums, int numsSize) { int tmp = 0; for(int i=1;i<numsSize;++i){ tmp = nums[0] & (tmp ^ nums[i]); nums[0] = tmp | (nums[0] ^ nums[i]); } return nums[0];}
Comment
This is not my idea,but it is so brilliant.Shamefully…I have not figured out it…
0 0
- 【latched】Single Number II
- latched之Single Number III
- Single Number & Single Number II
- Single Number & Single Number II
- Single Number II - leetcode
- Leetcode: Single Number II
- Single Number II
- Single Number II
- [LeetCode] Single Number II
- LeetCode: Single Number II
- leetcode -- Single Number II
- [leetcode]Single Number II
- [LeetCode] Single Number II
- 【leetcode】Single Number II
- Single Number I & II
- LeetCode:Single Number II
- Leetcode: Single Number II
- Single Number II
- js事件绑定的方法
- java多态的理解
- HDU 4104 Discount
- android中UDP编程的注意事项
- Socket 通信原理(Android客户端和服务器以TCP&&UDP方式互通)
- 【latched】Single Number II
- Oracle11g新特性影响EXP导出,ORA-01455的处理
- mybatis 校验报错问题[myeclipse 8.5]
- Android开发---音乐播放器(服务类的实现)
- ORA-00257 archiver error. 错误的处理方法
- 南邮 OJ 1383 Knights
- 基础知识(六)KD-Tree快速最近邻搜索
- UIScrollView那些事
- 深入理解HTTP协议