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LeetCode 476: Number Complement

Difficulty: Easy

Tags: Bit Manipulation

问题描述

给定一个正整数 num，输出其“补足数”。

补足数：num 的二进制表示翻转后再转为十进制数即为 num 的补足数。

解决方案

十进制 -> 二进制 -> 十进制。

▷ Time Complexity: O(lgN)

▷ Space Complexity: O(1)

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class Solution {
public:
    int findComplement(int num) {
        int array[32];
        int counter = 31;
        int head_position = 0;
        int num_length = 0;
        int complement_number = 0;
        
        for(int i=0; i<32; i++)
        {
            array[i] = 0;
        }
        
        // convert num to binary representation
        if(num == 1)    array[31] = 1;
        else if(num >= 2)
        {
            while(num/2 != 0)
            {
                if(num%2 == 1)    array[counter] = 1;
                num /= 2;
                counter--;
            }
            array[counter] = 1;
        }
    
        for(int i=0; i<32; i++)
        {
            if(array[i] == 1)
            {
                head_position = i;
                break;
            }
        }
    
        // get the complement number
        num_length = 32 - head_position;
        int complement[num_length];
    
        for(int i=0; i<num_length; i++)
        {
            complement[i] = 1 - array[head_position + i];
        }
    
        for(int i=0; i<num_length; i++)
        {
            complement_number += complement[num_length-i-1]*pow(2,i);
        }
        
        return complement_number;
    }
};

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