题目描述

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.

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.

解法一

思路

这题最容易想到的方法就是直接用~按位取反了,但这样有个明显的问题就是即使是前缀的0也会被取反,如int型5的二进制表示是0000 0000 0000 0000 0000 0000 0000 0101,其按位取反的结果是1111 1111 1111 1111 1111 1111 1111 1010,这样前面就多了29个1,我们还得想办法把这些1变成0,从而得到0000 0000 0000 0000 0000 0000 0000 0010,即十进制2。

但我们不妨换个思路,用异或^来求解,比如101 ^ 111 = 010。那么怎么得到111呢?考虑111 + 1 = 1000,而1000又是 最小的 大于101只有一位是1 的二进制数。

所以解决方法出来了:

  1. 找到最小的大于原数字的二进制值仅有一位为1的数;
  2. 将此数减1;
  3. 与原数字按位求异或。

Python

class Solution(object):
    def findComplement(self, num):
        """
        :type num: int
        :rtype: int
        """
        all_ones = 1
        while all_ones <= num:
            all_ones <<= 1
        all_ones -= 1
        return all_ones ^ num

Java

public class Solution {
    public int findComplement(int num) {
        long allOnes = 1;
        while (allOnes <= num) {
            allOnes <<= 1;
        }
        allOnes -= 1;
        return (int) allOnes ^ num;
    }
}

C++

class Solution {
public:
    int findComplement(int num) {
        long allOnes = 1;
        while (allOnes <= num) {
            allOnes <<= 1;
        }
        allOnes -= 1;
        return (int) allOnes ^ num;
    }
};