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.
求两个整数的二进制表示中,不同的 bit 位数量。异或操作符,相同位运算后为 0,不同位运算后为 1,因此只需要求两个整数经过异或之后,对应的二进制表示中 1 的数量。
class Solution {
public int hammingDistance(int x, int y) {
return hammingWeight(x^y);
}
public int hammingWeight(int n) {
int count = 0;
while (n != 0) {
count ++;
n &= (n - 1);
}
return count;
}
}
使用 Integer.bitCount 获取二进制数 bit 位中 1 的数量。
class Solution {
public int hammingDistance(int x, int y) {
return Integer.bitCount(x^y);
}
}