LeetCode[191] Number of 1 Bits

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.

依次移位

复杂度
O(N), O(1), N = number of bits in the interger

思路
依次移动位数进行计算。
The unsigned right shift operator ">>>" shifts a zero into the leftmost position, while the leftmost position after ">>" depends on sign extension.
So when the first bit is 1, if use >>, 1 will always be there, then the loop will never end.

代码

public int hammingWeight(int n) {
    int cnt = 0;
    while(n != 0) {
        if((n & 1) == 1) {
            cnt ++;
        }
        // must use unsigned operation
        n = n >>> 1;
    }
    return cnt;
}

利用性质 n & (n - 1)

复杂度
O(N), O(1)， N = number of 1 bits in the number

思路
consider n & (n - 1) always eliminates the least significant 1.

代码

public int hammingWeight(int n) {
    int cnt = 0;
    // loop times = number of 1's in the n
    while(n != 0) {
        n = n & (n - 1);
        cnt ++;
    }
    return cnt;
}