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;
};