leetcode 67. 二进制求和

给定两个二进制字符串，返回他们的和（用二进制表示）。

输入为非空字符串且只包含数字 1 和 0。

示例 1:

输入: a = "11", b = "1"
输出: "100"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

#include<algorithm>
class Solution {
public:
    string addBinary(string a, string b) {
        if(a.size()<b.size()){//让a保存a,b中最长的字符串
            string temp = a;
            a = b;
            b = temp;
        }//翻转字符串，便于运算
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());
        int carry = 0;
       for(int i=0; i<b.size(); i++){
           int t = (a[i]-'0') + (b[i]-'0') + carry;
           if(t>=2){//满2进1
               a[i] = (t%2)+'0';
               carry = 1;
           }else{
               a[i] = t + '0';
               carry = 0;
           }
       }
        for(int i=b.size(); i<a.size(); i++){
            int t = (a[i]-'0') + carry;
            if(t>=2){
                a[i] = '0';
                carry = 1;
            }else{
                a[i] = t + '0';
                carry = 0;
            }
        }
        if(carry == 1) a += '1';
        reverse(a.begin(), a.end());
        return a;
    }
};

在b的后面添加'0', 让加法趋于一致

#include<algorithm>
class Solution {
public:
    string addBinary(string a, string b) {
        if(a.size()<b.size()){//让a保存a,b中最长的字符串
            string temp = a;
            a = b;
            b = temp;
        }//翻转字符串，便于运算
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());
        int carry = 0;
        for(int i=b.size(); i<a.size(); i++) b += '0';
       for(int i=0; i<a.size(); i++){
           int t = (a[i]-'0') + (b[i]-'0') + carry;
           if(t>=2){//满2进1
               a[i] = (t%2)+'0';
               carry = 1;
           }else{
               a[i] = t + '0';
               carry = 0;
           }
       }
        if(carry == 1) a += '1';
        reverse(a.begin(), a.end());
        return a;
    }
};

进一步精简代码，把重复的操作合并，减少不必要的操作

#include<algorithm>
class Solution {
public:
    string addBinary(string a, string b) {
        if(a.size()<b.size()) swap(a, b);
        int carry = 0;
        for(int i=b.size(); i<a.size(); i++) b = '0' + b;
        for(int i=a.size()-1; i>=0; i--){
           int t = (a[i]-'0') + (b[i]-'0') + carry;
           a[i] = (t%2) + '0';
           carry = t/2;
       }
        if(carry == 1) a = '1' + a;
        return a;
    }
};