数的补数 Number Complement

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最新推荐文章于 2025-09-05 14:01:23 发布

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文章标签： python

原文链接：https://my.oschina.net/liyurong/blog/1518541

本文介绍了如何通过两种方法实现给定正整数的按位取反操作，即补码计算。第一种方法通过计算掩码并与原数进行位运算得到结果；第二种方法则利用了异或的性质来简化计算过程。

2019独角兽企业重金招聘Python工程师标准>>>

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:

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: 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);
}
}