Note:

1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
2. You could assume no leading zero bit in the integer’s binary representation.

Example 1:

Input: 5

Output: 2

Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.

Example 2:

Input: 1

Output: 0

Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.

idea: we only want to flip the lower k bits (k is the bits number of num)

Solution 1: use mask to find the leftmost 1 of num, and then use ~mask to make only the k lowest bits to be 1(k is the bits number of num), and then & ~num to get the result.

class Solution {
public:
    int findComplement(int num) {
    unsigned mask=INT_MAX;
    while((mask&num)!=0){
        mask= mask << 1;
    }
    return (~mask & ~num);
    }
};

 Solution 2: the needed mask is (2^(int)log(num))-1=(1<<(int)log(num))-1

class Solution {
public:
    int findComplement(int num) {
        return ~num & ((1 <<(int)log2(num))-1);
    }
};