LeetCode_461-hamming distance
2017-03-04 14:14
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Solution one：
思路：按位与直接比较
int hammingDistance(int x, int y) {
int c = 0;
for(int i = 32;i>=0;i--){
if((x & (1<<i)) != (y & (1<<i))){
c++;
}
}
return c;
}
Solution two：
思路：求出x与y的异或，再求出其二进制中的非零值
int hammingDistance(int x, int y) {
int c = 0,d=x^y;
while(d){
if(d%2)
{
c++;
}
d = d/2;
}
return c;
}
思路：按位与直接比较
int hammingDistance(int x, int y) {
int c = 0;
for(int i = 32;i>=0;i--){
if((x & (1<<i)) != (y & (1<<i))){
c++;
}
}
return c;
}
Solution two：
思路：求出x与y的异或，再求出其二进制中的非零值
int hammingDistance(int x, int y) {
int c = 0,d=x^y;
while(d){
if(d%2)
{
c++;
}
d = d/2;
}
return c;
}
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