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Formatted question description: https://leetcode.ca/all/476.html
476. Number Complement (Easy)
Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
Note:
- The given integer is guaranteed to fit within the range of a 32-bit signed integer.
- 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.
Companies:
Cloudera
Related Topics:
Bit Manipulation
Solution 1.
For example:
num = 00000101
mask = 11111000
~mask & ~num = 00000010
// OJ: https://leetcode.com/problems/number-complement/
// Time: O(1)
// Space: O(1)
class Solution {
public:
int findComplement(int num) {
unsigned mask = ~0;
while (num & mask) mask <<= 1;
return ~mask & ~num;
}
};
-
class Solution { public int findComplement(int num) { int power = (int) (Math.log(num) / Math.log(2)); long max = (long) Math.pow(2, power + 1) - 1; return (int) (max - num); } } ############ 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; } }
-
// OJ: https://leetcode.com/problems/number-complement/ // Time: O(1) // Space: O(1) class Solution { public: int findComplement(int num) { unsigned mask = ~0; while (num & mask) mask <<= 1; return ~mask & ~num; } };
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class Solution(object): def findComplement(self, num): i = 1 while i <= num: i = i << 1 return (i - 1) ^ num
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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 }