leetcode 67. Add Binary

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;    }};