LeetCode 461. Hamming Distance-Java
2016-12-27 19:59
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思路：
(1)异或运算：先求x^y的结果res。
(2)再依次求int32位的res每一位与1做&运算的结果，若不为0则Hamming Distance +1
(3)每判断完一位，res右移一位继续判断下一位
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public class Solution {
public int hammingDistance(int x, int y) {
int res = x ^ y;
int count = 0;
for (int i = 0; i < 32; i++) {
if ((res & 1) != 0)
count++;
res >>= 1;
}
return count;
}
}
Runtime：10ms
(1)异或运算：先求x^y的结果res。
(2)再依次求int32位的res每一位与1做&运算的结果，若不为0则Hamming Distance +1
(3)每判断完一位，res右移一位继续判断下一位
[html] view
plain copy
public class Solution {
public int hammingDistance(int x, int y) {
int res = x ^ y;
int count = 0;
for (int i = 0; i < 32; i++) {
if ((res & 1) != 0)
count++;
res >>= 1;
}
return count;
}
}
Runtime：10ms
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