【LeetCode】191.Number of 1 Bits

1 问题

Write a function that takes the binary representation of an unsigned integer and returns the number of ‘1’ bits it has (also known as the Hamming weight).

Note:

Note that in some languages, such as Java, there is no unsigned integer type. In this case, the input will be given as a signed integer type. It should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned.

In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 3, the input represents the signed integer. -3.

1.1 Example 1

Input: n = 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three ‘1’ bits.

1.2 Example 2

Input: n = 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one ‘1’ bit.

1.3 Example 3

Input: n = 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one ‘1’ bits.

说明

• The input must be a binary string of length 32.

2 解题思路

非常经典的题目。记得有本数据结构中的算法就是这样子解决的：n&(n-1). 只要循环下去，n!=0，就可以继续统计1的个数。

3 代码

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight2(int n) {
        int ans = 0;
        for (int i = 0; i < 32; i++) {
            ans += (n >> i & 1);
        }
        return ans;
    }

    public int hammingWeight(int n) {
        int ans = 0;
        while (n != 0) {
            n = n & (n - 1);
            ans++;
        }
        return ans;
    }
}