计算力扣平台上的汉明距离问题解析与优化算法

官方的解法

然后再配上这个代码就一目了然了

class Solution {public int totalHammingDistance(int[] nums) {int len&＃61;nums.length;int[] bitCount &＃61; new int[32];if(len <&＃61; 1){return 0;}for(int numIndex &＃61; 0; numIndex < len; numIndex&＃43;&＃43;){for(int bitIndex &＃61; 0; bitIndex < 32; bitIndex&＃43;&＃43;){bitCount[bitIndex] &＃43;&＃61; nums[numIndex] & 1;nums[numIndex] &＃61; nums[numIndex] >> 1;if(nums[numIndex] &＃61;&＃61; 0){break;}}}int oneCount &＃61; 0;for(int bitIndex &＃61; 0; bitIndex < 32; bitIndex&＃43;&＃43;){oneCount &＃43;&＃61; bitCount[bitIndex] * (len - bitCount[bitIndex]);}return oneCount;}
}


上边就是用bitCount [bitIndex] 从第0位累加到32位&＃xff0c;分别记录数组中每个数字对应的二进制位&＃xff0c;然后用官方提供的那种思维 t*(n-t)记录每一组的汉明距离&＃xff0c;最后再把32组累加起来就是总的汉明距离了

int hammingDistance(int x, int y){long long z &＃61; (x ^ y);int ret &＃61; 0;while (z !&＃61; 0) {z &＃61; (z & (z - 1));ret&＃43;&＃43;;}return ret;}int totalHammingDistance(vector<int>& nums){int ret &＃61; 0, n &＃61; nums.size();// 遍历所有的两两组合for (int i &＃61; 0; i < n - 1; i&＃43;&＃43;) {for (int j &＃61; i &＃43; 1; j < n; j&＃43;&＃43;) {ret &＃43;&＃61; hammingDistance(nums[i], nums[j]);}}return ret;}这是属于暴力求解的方法
再看看官方的解法![在这里插入图片描述](https://img-blog.csdnimg.cn/20200507232753581.png?x-oss-process&＃61;image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RlbGV0ZV9idWc&＃61;,size_16,color_FFFFFF,t_70)