191. Number of 1 Bits*

191. Number of 1 Bits*

https://leetcode.com/problems/number-of-1-bits/

题目描述

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

Example 1:

Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three '1' bits.

Example 2:

Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one '1' bit.

Example 3:

Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one '1' bits.

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, the input will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 3 above the input represents the signed integer -3.

Follow up:

If this function is called many times, how would you optimize it?

C++ 实现 1

做这道题需要了解一个结论: 思路 2: https://leetcode.com/problems/number-of-1-bits/discuss/55106/Python-2-solutions.-One-naive-solution-with-built-in-functions.-One-trick-with-bit-operation

2.Using bit operation to cancel a 1 in each round

Think of a number in binary n = XXXXXX1000, n - 1 is XXXXXX0111. n & (n - 1) will be XXXXXX0000 which is just cancel the last 1

也就是说, n & (n - 1) 的结果是 n 中最右边的 1 变为了 0, 那么持续这样操作下去, 最终 n 就会变为 0.

另外, 这个思路也可以用来处理 338. Counting Bits** 这道题.

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