leetcode解题: Number of 1 Bits (191)

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

For example, the 32-bit integer ’11’ has binary representation 00000000000000000000000000001011, so the function should return 3.

解法1：O(M) M is number of total set bits

运用n & (n -1)消掉最高位的set bit的思想，可以不停重复这个操作直到n变为0，
这样的复杂度可以减少为所有1的个数
C++

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class Solution {
public:
    int hammingWeight(uint32_t n) {
        if (n == 0) {
            return 0;
        }
        int count = 0;
        while (n != 0) {
            ++count;
            n = n & (n - 1);
        }
        return count;
    }
};

解法2：O(N) N is number of total bits

用右移的方法来判断每一位是否是1直到n变为0
C++

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class Solution {
public:
    int hammingWeight(uint32_t n) {
        int res = 0;
        while (n != 0) {
            res += n & 1;
            n =  n >> 1;
        }
        return res;
    }
};

Java

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