LeetCode 476. Number Complement （数的补数）

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.

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 题目标签：Bit Manipulation

　　这道题目给了我们一个int数，让我们找到它的补数。而且这个数字是正数，那么我们只要把除了leading zeros的数字都反转一下就可以了。设一个32次的for loop， 利用 & 1 来判断最后边的bit 是不是0， 如果是0，那么我们需要反转它，设一个x，规律是1，2，4，8。。。2进制， 如果是0，那么就加x，如果是1，那么不需要加。然后利用 >> 1来移动num 向右一位。 当num == 1的时候，意味着所有1左边的数字都是0了，这个时候已经不需要在继续反转下去了。

这道题目看上去挺简单的，但是一下子还没做出来。。。看来今天状态不佳，适合出去玩。

 

Java Solution:

Runtime beats 61.45%

完成日期：06/29/2017

关键词：Bit Manipulation

关键点：利用 & 1 判断最右边的bit； 利用 >> 来移动bits；判断num == 1 来终止反转

 public class Solution
 {
     public int findComplement(int num)
     {
         int res = 0;
         int x = 1;
 
         for(int i=0; i<32; i++)
         {
             if((num & 1) == 0) // meaning num's bit is 0, need to flip this bit.
                 res += x;
 
             x = x * 2; // increase the x 
 
             if(num == 1) // if current num is == 1, meaning all the leading numbers are zeros; stop
                 break;
 
             num = num >> 1; // shift num to right by 1 bit.
 
         }
 
         return res;
     }
 }

参考资料：

http://www.cnblogs.com/grandyang/p/6275742.html

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