题目描述：
Given a positiveinteger, output its complement number. The complement strategy is to flip thebits of its binary representation.
Note:
The given integer isguaranteed to fit within the range of a 32-bit signed integer.
You could assume noleading zero bit in the integer’s binary representation.
Example 1:
Input: 5
Output: 2
Explanation: Thebinary representation of 5 is 101 (no leading zero bits), and its complement is010. So you need to output 2.
Example 2:
Input: 1
Output: 0
Explanation: Thebinary representation of 1 is 1 (no leading zero bits), and its complement is0. So you need to output 0.
题目理解：
给定一个正整数，输出它的补码。补码是将该正整数的二进制全部翻转后得到的整数
我的分析：
1. 依次，用2除，求余数，求商；
2. 由此得到二进制的反序列；
3. 将序列放入整型数组，再依次取出翻转求出补码。
我的解答：
public static int findComplement(int num){ int result = 0; int [] bitnum = new int[32]; int range = 0; for(int i = 0; num != 0; i++){ bitnum[i] = num % 2; num /= 2; range = i; } for(int i = 0; i <= range; i++){ if(bitnum[i] == 0) result += (bitnum[i] + 1) * Math.pow(2, i); else result += (bitnum[i] - 1) * Math.pow(2, i); } return result; } 其他的解答1： //100110, its complement is 011001, the sum is 111111. So we only need get the min number large or equal to num, then do substraction public int findComplement(int num) { int i = 0; int j = 0; while (i < num) { i += Math.pow(2, j); j++; } return i - num; }对解答1的改动：
一个数乘2，相当于左移一位
//Same idea, but using bit manipulation instead of Math.pow(). public class Solution { public int findComplement(int num) { int n = 0; while (n < num) { n = (n << 1) | 1; } return n - num; } } 更快的解答1： import java.lang.*; public class Solution { public int findComplement(int num) { int counter = (int) (Math.floor((Math.log(num)/Math.log(2))) + 1); return (((~num) << (32-counter)) >>> (32-counter)); } }更快的解答2：
public class Solution { public int findComplement(int num) { return num ^ ((Integer.highestOneBit(num) << 1) - 1); } }