single-number-ii
来源:互联网 发布:已知矩阵怎么求行列式 编辑:程序博客网 时间:2024/06/08 00:17
题目描述
Given an array of integers, every element appears three times except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
class Solution {public: int singleNumber(int A[], int n) { int one=0; //记录某一位,1只出现一次时为1. int two=0;//记录某一位,1只出现两次时为1; int three;//记录某一位,1只出现3次时为1; for(int i=0;i<n;++i) { two|=one&A[i]; //如果1只出现了一次,并且这次又为1,则two为1,如果1上次出现,这次没出现则two保持原来的状态。 one^=A[i]; //如果1又出现了,证明不是只出现1次,要置0.之后重新统计 three=one&two;//只出现一次,并且又出现了一次,证明出现了3次。 one&=~three;//将出现3次1的位置置零 two&=~three; } return one; }};
阅读全文
0 0
- 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
- leetcode :Single Number II
- Leetcode Single Number II
- Python | 数据整合
- CSS布局初体验
- ABAP 中的Screen, Dynpro, PBO, PAI小结
- day15 集合(一)
- 循环
- single-number-ii
- Unity3D之奇葩代码(1)
- 详解DTD & 渗透解析Doctype
- 数学知识
- SpringMVC结合POI复杂报表打印
- 《用户体验的要素-以用户为中心的Web设计》——小结
- 调试
- 文章标题
- IO多路复用之epoll总结