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题目
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.
分析
题目含义是求两个整数对应的二进制串中不同比特位值的个数。
可以转化为 求 (x^y)对应二进制串中1的个数。
代码
class Solution {
public int hammingDistance(int x, int y) {
return countNumOf1Bits(x ^ y);
}
public int countNumOf1Bits(int n) {
int count = 0;
while (n > 0) {
if ((n & 1) == 1) {
++count;
}
n >>= 1;
}
return count;
}
}