LeetCode------------single-number-ii
来源:互联网 发布:linux 日志切割脚本 编辑:程序博客网 时间:2024/06/03 23:38
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?
用map的话与上一道的答案可以完全一样。
2
3
4
5
6
7
8
9
10
11
12
13
14
class
Solution {
public
:
int
singleNumber(
int
A[],
int
n) {
map<
int
,
int
> mp;
for
(
int
i=
0
; i<n; i++){
mp[A[i]]++;
}
int
j=
0
;
while
(mp[A[j]] !=
1
){
j++;
}
return
A[j];
}
};
Single Number的本质,就是用一个数记录每个bit出现的次数,如果一个bit出现两次就归0,这种运算采用二进制底下的位操作^是很自然的。Single Number II中,如果能定义三进制底下的某种位操作,也可以达到相同的效果,Single Number II中想要记录每个bit出现的次数,一个数搞不定就加两个数,用ones来记录只出现过一次的bits,用twos来记录只出现过两次的bits,ones&twos实际上就记录了出现过三次的bits,这时候我们来模拟进行出现3次就抵消为0的操作,抹去ones和twos中都为1的bits。
1
2
3
4
5
6
7
8
9
10
11
12
13
14
public
int
singleNumber(
int
[] A) {
int
ones =
0
;
//记录只出现过1次的bits
int
twos =
0
;
//记录只出现过2次的bits
int
threes;
for
(
int
i =
0
; i < A.length; i++){
int
t = A[i];
twos |= ones&t;
//要在更新ones前面更新twos
ones ^= t;
threes = ones&twos;
//ones和twos中都为1即出现了3次
ones &= ~threes;
//抹去出现了3次的bits
twos &= ~threes;
}
return
ones;
}
0 0
- Single Number II - leetcode
- Leetcode: 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
- LeetCode: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
- leetcode Single Number II
- [LeetCode],Single Number II
- LeetCode | Single Number II
- 千里码4:GET&POST
- 35. 内容协商和转码
- 面向对象_final修饰变量的初始化时机
- Struts2学习——0100HelloStruts
- 基础算法模块总结
- LeetCode------------single-number-ii
- noip2013积木大赛
- Oracle 的过程和函数
- hdu3526(最小费用流)
- 异步任务下载网络图片
- 236. Lowest Common Ancestor of a Binary Tree
- ViewPager详解(一)简单介绍
- AngularJS: $broadcast $emit $on
- [LeetCode-Java]35. Search Insert Position