476. Number Complement

Posted 2020-08-29

题目

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.

 

分析

将int的二进制表示中所有0变为1，1变为0

解答

解法1：（我）

例1：

          5: 00000000000000000000000000000101

        ~5: 11111111111111111111111111111010 (补码形式，原码是10000000000000000000000000000110，值为-6)

        -2³: 11111111111111111111111111111000 (补码形式，原码是10000000000000000000000000001000，值为-2³)

~5-(-2³): 00000000000000000000000000000010 (值为2)

 

---

 

   int的最大值：01111111111111111111111111111111          2³¹-1

int的非最小值：11111111111111111111111111111111       -(2³¹-1)

   int的最小值：10000000000000000000000000000000        -2³¹ （特殊：使用以前的-0的补码来表示, 所以-2³¹并没有原码和反码表示）

例2：

               2³¹-1: 01111111111111111111111111111111

          ~(2³¹-1): 10000000000000000000000000000000 (值为-2³¹)

                 -2³¹: 10000000000000000000000000000000

~(2³¹-1)-(-2³¹): 00000000000000000000000000000000 (值为0)

注：此处~(2³¹-1)-(-2³¹)不能写为~(2³¹-1)+2³¹，因为int中正数2³¹超出最大值范围，会被解析成2³¹-1，而负数(-2³¹)没有超出最小值范围

1 public class Solution {
2     public int findComplement(int num) {
3         return ~num-(int)-Math.pow(2,32-Integer.numberOfLeadingZeros(num));
4     }
5 }