数的补数 Number Complement

问题：

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.（输出每个数的补码，实际上根据示例是要求实现按位取反）

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.

解决：

①  注意对输入的数值转换时是32位二进制数，高位为0，不算在计算范围内，应该跳过，从第一个非0数据开始。

class Solution { // 10ms
    public int findComplement(int num) {
        int tmp = (Integer.highestOneBit(num) << 1) - 1; //00..11..1
        num = ~ num;//111...取反之后的值
        return num & tmp;//000...补码
    }
}

进化版：异或的性质---与0相^保留原值，与1相^按位取反，与自身相^结果为0.

public class Solution { // 11ms
    public int findComplement(int num) {
        int mask = (Integer.highestOneBit(num) << 1) - 1;
        return num ^ mask;
    }
}

② 利用异或的性质。

class Solution { // 11ms
    public int findComplement(int num) {
        int tmp = num;
        int count = 0;//记录非0的位数
        while(tmp != 0){
            tmp /= 2;
            count ++;
        }
        return num ^ (int)(Math.pow(2,count) - 1);
    }
}

来源：oschina

链接：https://my.oschina.net/u/2968041/blog/1518541