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【LeetCode】190 & 191 - Reverse Bits & Number of 1 Bits

2015-08-04 00:06 597 查看
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 };
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