📃 题目描述

题目链接：剑指 Offer 15. 二进制中1的个数 (opens new window)

🔔 解题思路

循环检查二进制位

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;

        for (int i = 0; i < 32; i ++) {
            if (((n >> i) & 1) == 1) {
                count ++;
            }
        }

        return count;
    }
}

时间复杂度 O(N)

Brian Kernighan 算法

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        int count = 0;

        while (n != 0) {
            // 最后一位 1 变成 0
            n &= (n - 1);
            count ++;
        }

        return count;
    }
}

时间复杂度 O(LogN)

💥 复杂度分析

• 空间复杂度：
• 时间复杂度：