本文共 988 字,大约阅读时间需要 3 分钟。
1 2 3 4 5 6 7 8 9 10 | 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: 5Output: 2Explanation: 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: 1Output: 0Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0. |
题意:给一个正整数,返回其补数。所谓补数就是。。。。1的补数是0.把每一位求补返回就好。
1 2 3 4 5 6 7 8 9 10 11 | public class Solution { public int findComplement( int num) { int cp=num; int ans= 0 ; while (cp!= 0 ){ cp=cp/ 2 ; ans++; } return num^( int )(Math.pow( 2 ,ans)- 1 ); } } |
PS:利用了异或的性质
A^1=-A。
先用ans求出正整数的二进制位数,然后直接与项ans个1表示的二进制和其异或即可。
本文转自 努力的C 51CTO博客,原文链接:http://blog.51cto.com/fulin0532/1890949
转载地址:http://rewua.baihongyu.com/