leetcode 190. Reverse Bits 191. Number of 1 Bits

最新推荐文章于 2023-11-19 19:00:00 发布

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最新推荐文章于 2023-11-19 19:00:00 发布

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本文介绍了两种实用的位操作技巧：一是反转一个32位无符号整数的位；二是计算一个32位无符号整数中1的个数（汉明重量）。通过具体的C++代码实现展示了如何高效地完成这两种操作。

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190. Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

Follow up:

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

class Solution {
public:
    uint32_t reverseBits(uint32_t n)
    {
        uint32_t ret = n;
        for (int i = 0; i < 16; i ++)
        {
            int left = 31 - i, right = i;
            if ((((n>>left) & 1)  ^ ((n>>right) & 1)) == 0)  //说明要换的那两个位置相同，不用互换
                continue;
            else
            {
                if (((n>>right) & 1) == 1)  //低位是0，高位是1.然后交换为高位1，低位0
                {
                    ret = ret & ~(1<<right);    //设置低位为0
                    ret = ret | (1<<left);      //设置高位为1
                } else {
                    ret = ret & ~(1<<left);    //设置高位为0
                    ret = ret | (1<<right);    //设置低位为1
                }
            }
        }
        return ret;
    }
};

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.

class Solution {
public:
    int hammingWeight(uint32_t n) 
    {
        int ret = 0;
        for (int i = 0; i < 32; i++)
            ret += ( ((n>>i) & 1) == 1) ? 1 : 0;
        return ret;
    }
};