LeetCode_461. Hamming Distance

/*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.*/

题目的意思是：找到x和y的二进制位中，不相同的个数；

自己的思路的先将x，y转化为二进制存储，然后比较，结果超时了：

var hammingDistance = function(x, y) {    var MAX = Math.pow(2, 32);  if (x < 0 || y < 0 || x > MAX || y > MAX) {        console.log("max");        return false    }    var arr1 = revertTobit(x);    var arr2 = revertTobit(y);    while (arr1.length != arr2.length) {        if (arr1.length < arr2.length) {            arr1.unshift(0);        } else {            arr2.unshift(0)        }    }    var count = 0;    for (var i = 0; i < arr1.length; i++) {        if (arr1[i] != arr2[i]) {            count++;        }    }    return count;};/*将一个数转换为二进制数组存储*/var revertTobit = function(x) {    var arr = [];    while (x != 1) {        var bit = x % 2;        x = Math.floor(x / 2);        arr.unshift(bit);    }    arr.unshift(1);    return arr;}

正确的方法是采用位运算，x异或y 得到的z

关键在于计算z有多少个1，方法：
用z和z-1与运算；这样做的与运算每次都把最右边的1变为0，最后直到z为0；

var hammingDistance = function(x, y) {  var count = 0;  var n = x ^ y;  while (n) {    ++count;    n = (n - 1) & n;  }  return count;};