leetcode-461-Hamming Distance

题目描述：

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4Output: 2Explanation:
1   (0 0 0 1)
4   (0 1 0 0)↑   ↑The above arrows point to positions where the corresponding bits are different.

要完成的函数：

int hammingDistance(int x, int y)

说明：

给定两个32位的有符号型整数，当每一位不相同时，加上1，输出最后一共有多少位不相同。

代码如下：

    int hammingDistance(int x, int y) {int xx,yy,count=0;for(int i=0;i<32;i++){xx=x&1;yy=y&1;if(xx!=yy)count++;x>>=1;y>>=1;}return count;}

代码浅显易懂，实测6ms，beats 72.58% of cpp submissions。

这道题在评论区还看到了令人耳目一新的做法，采用异或，当两位相同时输出0，当两位不同时输出1。不过这种做法的时间花费实际还是跟上述代码差不多的。

代码如下：

    int hammingDistance(int x, int y) {int result=x^y,t1,count=0;for(int i=0;i<32;i++){t1=result&1;if(t1==1)count++;result>>=1;}return count;}

转载于:https://www.cnblogs.com/chenjx85/p/8948404.html

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leetcode-461-Hamming Distance

题目描述：

要完成的函数：

说明：

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