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.
含义:求两个数字的汉明距离
思路:先进行按位易或运算,不同的位结果为1,然后计算1个个数
1 public int hammingDistance(int x, int y) { 2 // return Integer.bitCount(x^y); 3 int res = 0, exc = x ^ y; 4 // for (int i = 0; i < 32; ++i) { 5 // res += (exc >> i) & 1; 6 // } 7 8 while (exc!=0) { 9 ++res; 10 exc &= (exc - 1); //消除最后一位1 11 } 12 return res; 13 }