leetcode 461. Hamming Distance
2017-04-17 23:18
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一.本题熟悉一个算法。 判断一个二进制x有几个1
即x&1 看x的最低位是否为1,然后将x右移再&1即可。知道x为0;
二.代码如下:
class Solution {
public:
int hammingDistance(int x, int y) {
int count=0;
int result = x^y;
while(result>0)
{
if((result&1)==1)
count++;
result>>=1;
}
return count;
}
};
即x&1 看x的最低位是否为1,然后将x右移再&1即可。知道x为0;
二.代码如下:
class Solution {
public:
int hammingDistance(int x, int y) {
int count=0;
int result = x^y;
while(result>0)
{
if((result&1)==1)
count++;
result>>=1;
}
return count;
}
};
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