【LeetCode】190 & 191

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 as00111001011110000010100101000000).

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

Related problem: Reverse Integer

Solution 1:

 class Solution {
 public:
    uint32_t reverseBits(uint32_t n) {      //runtime:4ms
        unsigned ret = 0;
        unsigned x = 1 << 31;
        unsigned y = 1;
        while(x){
            if(x & n)ret |= y;   //或ret += y;
            x >>= 1;
            y <<= 1;
         }
         return ret;
     }
 };

Solution 2:

 class Solution {
 public:
    uint32_t reverseBits(uint32_t n) {      //runtime:4ms
        unsigned int bit = 0;
        unsigned int result = 0;
        while(bit<32)
        {
            if((n>>bit) & 1 == 1)
                result = result + (1<<(31-bit));
            bit ++;
        }
    
        return result;
    }
 };

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.

Solution：n&(n-1)实现n与n-1的按位与，消除最后一位1

 1 class Solution {
 2 public:
 3     int hammingWeight(uint32_t n) {
 4         int count=0;
 5         while(n){
 6             n &= (n-1);
 7             count++;
 8         }
 9         return count;
10     }
11 };