leetcode-461-Hamming Distance
2016-12-29 19:31
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问题
题目:[leetcode-461]思路
由于题目限制了整数范围,所以肯定是4个字节。那么内存中表示就是4个字节。所以,把十进制数字转换为二进制之后,进行比较即可。
代码
class Solution { public: int hammingDistance(int x, int y) { std::string left = decimal2binary(x); std::string right = decimal2binary(y); int diff = 0; for( int i = 0; i < 32; ++i ) { if( left[i] != right[i] ) ++diff; } return diff; } private: std::string decimal2binary(int val){ std::string ret(32, '0'); int idx = 0; while(val){ ret[idx] = '0' + (val%2); val /= 2; ++idx; } return ret; } };
思路1
也没有本质上简单很多的算法,做了一点修正。代码1
class Solution { public: int hammingDistance(int x, int y) { return count_one( x ^ y ); } private: int count_one( int val ){ int ret = 0; while( val ){ if(val%2) ++ ret; val /= 2; } return ret; } };
思路2
和上面的办法一样。枚举每一位,但是这么写能快点。代码2
class Solution { public: int hammingDistance(int x, int y) { int tmp = x^y; int ans = 0; for(int i = 0; i < 32; ++i){ ans += (tmp >> i) & 1; } return ans; } };
思路3
下面这么做是最快的。num&(num-1)可以快速去除最右边的1。证明:不妨假设最右边一位1在n位,倒数第二位1在m位。
1xxxx1xxxx,对于这种情形。num-1得到
1xxxx01111,
所以num & (num-1)得到 1xxxxxxxxx
假设最低位1不在第0位,num-1之后,包括最低位1和后面的位全部变成相反数,与操作变成0。
假设最低位1在第0位,num-1之后,该位为0。与操作之后变成0;
代码3
class Solution { public: int hammingDistance(int x, int y) { int tmp = x^y; int ans = 0; while(tmp){ ++ans; tmp &= tmp-1; } return ans; } };
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