1009. Complement of Base 10 Integer

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本文详细解析了LeetCode上的题目1009：十进制整数的补码，介绍了如何将一个非负整数的二进制表示取反，并将其转换回十进制表示的过程。通过具体的示例和算法实现，帮助读者理解并掌握这一技巧。

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1009. Complement of Base 10 Integer

这是一道LeetCode上面的Easy题目，原文连接：https://leetcode.com/problems/complement-of-base-10-integer/

原文描述

Every non-negative integer N has a binary representation.  For example, 5 can be represented as "101" in binary, 11 as "1011" in binary, and so on.  Note that except for N = 0, there are no leading zeroes in any binary representation.

The complement of a binary representation is the number in binary you get when changing every 1 to a 0 and 0 to a 1.  For example, the complement of "101" in binary is "010" in binary.

For a given number N in base-10, return the complement of it's binary representation as a base-10 integer.

Example 1:
Input: 5
Output: 2
Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.
Example 2:

Input: 7
Output: 0
Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.
Example 3:

Input: 10
Output: 5
Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.
 

Note:

0 <= N < 10^9

该题是说把一个整数(0~10^9)的二进制取反得到的二进制，返回二进制对应的十进制

思路

先把该整数的二进制数的每一位取出来，进行取反，再把每一位二进制数转换为十进制数相加就能得到结果。由于在Java进行取反的时候会出现诡异的事件，具体还不知道什么原因。（有知道的大佬可以留言说明，感谢！）所以采用了另外类似的方法：先把该整数的二进制数的每一位取出来，跟1进行异或，再把每一位二进制数转换为十进制数相加就能得到结果。
取反：把二进制数由0变成1，由1变成0
异或：如果两个二进制数相同则为0，不相同则为1

取反说明

	public static void main(String[] args) {
		int n = 10;
		System.out.println(n+"取反后的结果："+~n);
		int b = 0;
		while(n>0){
			b = ~(n&1);
			System.out.println((n&1)+"取反的结果："+b);
            n = n>>1;
		}
	}

程序执行结果：

10取反后的结果：-11
0取反的结果：-1
1取反的结果：-2
0取反的结果：-1
1取反的结果：-2

算法实现

class Solution {
    public int bitwiseComplement(int N) {
        if(N == 0)return 1;
        if(N == 1)return 0;
        double num = 0;
        int b = 0;
        int i = 0;
        while(N>0){
            b = (N&1)^1;
            num = num + b*Math.pow(2,i);
            i++;
            N = N>>1;
        }
        return (int)num;
    }
}