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.
方法1:
思路:不相同的位可以用异或运算标记出来,方法主要区别于如何统计异或结果中1的个数。这个方法先进行异或,然后用1 << i & x来统计1的次数
易错点:
- 1 << i 和 i >> 1意义不一样。还有 i << 1。 a << b:a向左shift b次
- & 是位与,&&是logic and:后者在这里是错误的
class Solution {
public:
int hammingDistance(int x, int y) {
int bit = x ^ y;
int result = 0;
for (int i = 0; i < 32; i++){
if ((1 << i) & bit){
result ++;
}
}
return result;
}
};
grandyang: http://www.cnblogs.com/grandyang/p/6201215.html