Problem

Count how many 1 in binary representation of a 32-bit integer.

Example

Given 32, return 1

Given 5, return 2

Given 1023, return 9

Challenge

If the integer is n bits with m bits. Can you do it in O(m) time?

Note

这道题，Olivia给我解决了两个疑问，还剩一个。首先是要用无符号右移运算符>>>，其次是可以用一个不断左移的二进制1作为参照。那么如何获得一个用1来进行补位的左移的1呢？

第一种解法，num右移，每次对末位进行比较，直到num为0；第二种解法，1左移，每次和num的第i位比较，直到i = 32；第三种解法，num和num-1逐位与，去1，直到num为0。

Solution

1.

public class Solution {    public int countOnes(int num) {        int count = 0;        while (num != 0) {            count += (num & 1);            num >>>= 1;        }        return count;    }};

2.

public class Solution {    public int countOnes(int num) {        int count = 0;        for(int i = 0 ; i < 32; i++) {            if((num & (1<

3.

public class Solution {    public int countOnes(int num) {        int count = 0;        while (num != 0) {            num &= num - 1;            count++;        }        return count;    }}

// 1111 1110 1110

// 1110 1101 1100// 1100 1011 1000// 1000 0111 0000