Given a positive integer num
, output its complement number. The complement strategy is to flip the bits of its binary representation.
Example 1:
Input: num = 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: num = 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.
Constraints:
- The given integer
num
is guaranteed to fit within the range of a 32-bit signed integer. num >= 1
- You could assume no leading zero bit in the integer’s binary representation.
- This question is the same as 1009: https://leetcode.com/problems/complement-of-base-10-integer/
class Solution:
def findComplement(self, num: int) -> int:
ans = 0
find = False
for i in range(30, -1, -1):
b = num & (1 << i)
if not find and b == 0:
continue
find = True
if b == 0:
ans |= (1 << i)
return ans
class Solution {
public int findComplement(int num) {
int ans = 0;
boolean find = false;
for (int i = 30; i >= 0; --i) {
int b = num & (1 << i);
if (!find && b == 0) {
continue;
}
find = true;
if (b == 0) {
ans |= (1 << i);
}
}
return ans;
}
}
class Solution {
public:
int findComplement(int num) {
int full = pow(2, int(log2(num)) + 1) - 1;
return full ^ num;
}
};
class Solution {
public:
int findComplement(int num) {
int ans = 0;
bool find = false;
for (int i = 30; i >= 0; --i)
{
int b = num & (1 << i);
if (!find && b == 0) continue;
find = true;
if (b == 0) ans |= (1 << i);
}
return ans;
}
};
func findComplement(num int) int {
ans := 0
find := false
for i := 30; i >= 0; i-- {
b := num & (1 << i)
if !find && b == 0 {
continue
}
find = true
if b == 0 {
ans |= (1 << i)
}
}
return ans
}