leetcode 67. Add Binary

拼搏现实的明天。 2022-06-09 11:46 78阅读 0赞

Given two binary strings, return their sum (also a binary string).

For example,  
a = “11”  
b = “1”  
Return “100”.

和上一道题一样，不过这里是按照二进制的加法。

代码如下：

public class Solution {
        
        public String addBinary(String a, String b)
        {
            a=a.trim();
            b=b.trim();
            if(a.length()<=0)
                return b;
            if(b.length()<=0)
                return a;
    
            int lena=a.length();
            int lenb=b.length();
            if(lena>lenb)
            {
                String tt=a;
                a=b;
                b=tt;
            }
            lena=a.length();
            lenb=b.length();
    
            String res="";
            int jinwei=0,i=lena-1,j=lenb-1;
            for(i=lena-1;i>=0;i--)
            {
                int sum=Integer.parseInt(a.charAt(i)+"")+Integer.parseInt(b.charAt(j)+"")+jinwei;
                j--;
                res=(sum%2)+res;
                jinwei=sum/2;   
            }
    
            for(i=j;i>=0;i--)
            {
                int sum=Integer.parseInt(b.charAt(i)+"")+jinwei;
                res=(sum%2)+res;
                jinwei=sum/2;
            }
    
            if(jinwei<=0)
                return res;
            else
            {
                res=jinwei+res;
                return res;
            }
        }
    }

C++的答案如下，和上一道题一样，很简单

代码如下：

#include <iostream>
    #include <sstream>
    
    using namespace std;
    
    class Solution 
    {
        
    public:
        string addBinary(string a, string b) 
        {
            string res = "";
            int i = a.length() - 1 , j = b.length() - 1;
            int jinwei = 0;
            while (i >= 0 && j >= 0)
            {
                int x = (int)(a[i--] - '0');
                int y = (int)(b[j--] - '0');
                int sum = x + y + jinwei;
                jinwei = sum / 2;
                res = to_string(sum % 2) + res;
            }
    
            while (i >= 0)
            {
                int x = (int)(a[i--] - '0');
                int sum = x + jinwei;
                jinwei = sum / 2;
                res = to_string(sum % 2) + res;
            }
    
            while (j >= 0)
            {
                int y = (int)(b[j--] - '0');
                int sum = y + jinwei;
                jinwei = sum / 2;
                res = to_string(sum % 2) + res;
            }
    
            if(jinwei>0)
                res = to_string(jinwei) + res;
            return res;
        }
    };