[LeetCode] Hamming Distance 汉明距离

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 ≤ xy < 231.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
↑   ↑

The above arrows point to positions where the corresponding bits are different.

class Solution {
public:
int hammingDistance(int x, int y) {
int res = 0;
for (int i = 0; i < 32; ++i) {
if ((x & (1 << i)) ^ (y & (1 << i))) {
++res;
}
}
return res;
}
};

class Solution {
public:
int hammingDistance(int x, int y) {
int res = 0, exc = x ^ y;
for (int i = 0; i < 32; ++i) {
res += (exc >> i) & 1;
}
return res;
}
};

class Solution {
public:
int hammingDistance(int x, int y) {
int res = 0, exc = x ^ y;
while (exc) {
++res;
exc &= (exc - 1);
}
return res;
}
};

class Solution {
public:
int hammingDistance(int x, int y) {
if ((x ^ y) == 0) return 0;
return (x ^ y) % 2 + hammingDistance(x / 2, y / 2);
}
};

https://discuss.leetcode.com/topic/72089/java-3-line-solution

https://discuss.leetcode.com/topic/72093/java-1-line-solution-d

https://discuss.leetcode.com/topic/72289/0ms-c-two-line-solution

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-12-20 01:05 Grandyang 阅读(...) 评论(...) 编辑 收藏