LeetCode基础--位运算(in progress)

Bitwise XOR

XOR(^) ：异或，数学符号为 ^

1^0 = 1  0^1 = 1  0^0 = 01^1 = 0  

XOR by 1 可以像开发一样从 1 到 0 或者 从 0 到 1.
还有一个有意思的：

x^0 = x x^x = 0

应用 1：翻转数字 N 的所有二进制位。
解决方案：如果 N 是32位数字，则：

N ^((1 << 32) - 1 )

应用 2：不用运算符交换两个数字。
解决方案：

A = A ^ B;B = A ^ B;A = A ^ B;

Bitwise AND

1 & 1 = 11 & 0 = 00 & 1 = 00 & 0 = 0

Bitwise Inclusive OR

1 | 1 = 11 | 0 = 10 | 1 = 10 | 0 = 0

Bitwise Complement

https://xijunlee.github.io/2017/04/01/efficiently/
https://discuss.leetcode.com/topic/50315/a-summary-how-to-use-bit-manipulation-to-solve-problems-easily-and-efficiently
http://blog.csdn.net/fanoluo/article/details/49456275
https://www.hackerrank.com/topics/bit-manipulation