461.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, return the Hamming distance between them.
Example 1:
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| 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.
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Example 2:
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| Input: x = 3, y = 1
Output: 1
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位运算：
时间复杂度：O(logC)，本题中logC = log2^31
空间复杂度：O(1)
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| class Solution {
public:
int hammingDistance(int x, int y) {
int ans = 0;
while(x > 0 || y > 0) {
if((x & 1) != (y & 1))ans++;
x >>= 1;
y >>= 1;
}
return ans;
}
};
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| class Solution {
public:
int hammingDistance(int x, int y) {
int s = x ^ y, ret = 0;
while (s) {
ret += s & 1;
s >>= 1;
}
return ret;
}
};
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内置位运算：
时间复杂度：O(1)
空间复杂度：O(1)
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| class Solution {
public:
int hammingDistance(int x, int y) {
return __builtin_popcount(x ^ y);
}
};
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Brian Kernighan 算法：
时间复杂度：O(logC)，本题中logC = log2^31
空间复杂度：O(1)
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| class Solution {
public:
int hammingDistance(int x, int y) {
int s = x ^ y, ret = 0;
while (s) {
s &= s - 1;
ret++;
}
return ret;
}
};
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