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:
1 class Solution { 2 public: 3 uint32_t reverseBits(uint32_t n) { //runtime:4ms 4 unsigned ret = 0; 5 unsigned x = 1 << 31; 6 unsigned y = 1; 7 while(x){ 8 if(x & n)ret |= y; //或ret += y; 9 x >>= 1; 10 y <<= 1; 11 } 12 return ret; 13 } 14 };
Solution 2:
1 class Solution { 2 public: 3 uint32_t reverseBits(uint32_t n) { //runtime:4ms 4 unsigned int bit = 0; 5 unsigned int result = 0; 6 while(bit<32) 7 { 8 if((n>>bit) & 1 == 1) 9 result = result + (1<<(31-bit)); 10 bit ++; 11 } 12 13 return result; 14 } 15 };
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 };