191. Number of 1 Bits

题目：

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.

链接：  http://leetcode.com/problems/number-of-1-bits/

题解：

跟上题一样，应该有些很厉害的解法。自己只做了最naive的一种。

Time Complexity - O(1)， Space Complexity - O(1)

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;
    }
}

Brian Kernighan的方法

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

        for (count = 0; n != 0; count++) {
            n &= n - 1;     // clear the least significant bit set
        }

        return count;
    }
}

二刷：

Java:

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;
    }
}

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

三刷:

使用一个while 循环，每次把n的最低的非0位清零。这样比计算全32位计算的位数少，但也多了每次按位与&操作的比较。

Java:

Time Complexity - O(1)， Space Complexity - O(1)

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

Reference:

http://www.hackersdelight.org/hdcodetxt/pop.c.txt

https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetNaive