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姚屹晨

leetcode - [191] Number of 1 Bits

2020/06/23

问题描述

计算整数二进制形式中1的个数。

举个例子：

Example 1:
Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:
Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:
Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Update at 2020_0625

注意，n可能是负数，所以while循环中的条件不能是n > 0，

而应该是n != 0。

Update at 2020_0717

Thanks Y总的lowBit()

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int ans = 0;
        while (n != 0){
            n -= lowBit(n);
            ans++;
        }
        return ans;
    }
    
    private int lowBit(int x){
        return x & -x;
    }
}

解法1

直接来看下实现：

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        
        int ans = 0;
        
        while (n != 0){
            ans += (n & 1);
            n >>>= 1;
        }
        
        return ans;
    }
}

Enjoy it !


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