第56天。

刷道水题Hamming Distance:

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 = 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.

所谓的humming distance就是两个数在bit位上不同的个数，就int来说，最多就是全部不相同，也就是每个bit位都不一样，即humming distance.

我们可以利用异或来很快的求出来，异或可以让bit位不相同时置1，相同时置0.则两数异或后所得到的数中有bit位中1的个数就是humming distance:

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int hammingDistance(int x, int y) {
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    int t = x^y;
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    int ret = 0;
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    while(t!=0) {
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        ret += (t&1);
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        t >>= 1;
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    }
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    return ret;
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}