[LeetCode]--191. Number of 1 Bits

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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

先科普一个按位与的问题：

    System.out.println("---" + (1100010011 & 1));

这个输出是 1

    System.out.println("---" + (1100010010 & 1));

这个输出是 0

    System.out.println("---" + (1100010011 & 11));

这个输出是 11

以此类推。

我写的这个很好理解。每次按照无符号数右移一位然后赋值给自己，最后一位为1，res就会加一，为0就加零，最后返回即可。

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

这个是别人的答案，也是Accept的，但是我没看太懂。

public int hammingWeight2(int n) {
        if (n == 0)
            return 0;
        int count = 1;
        while ((n & (n - 1)) != 0) {
            n &= n - 1;
            count++;
        }
        return count;
    }

给出官方做法。

Approach #1 (Loop and Flip) [Accepted]

public int hammingWeight(int n) {
    int bits = 0;
    int mask = 1;
    for (int i = 0; i < 32; i++) {
        if ((n & mask) != 0) {
            bits++;
        }
        mask <<= 1;
    }
    return bits;
}

Approach #2 (Bit Manipulation Trick) [Accepted]

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

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