微软面试题： LeetCode 415. 字符串相加 出现次数：2

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 class Solution {
 5 public:
 6     string addStrings(string num1, string num2)
 7     {
 8         string res_str = "";
 9         int i = num1.size() - 1;
10         int j = num2.size() - 1;
11         int mv_bit = 0;
12         int sum = 0;
13         char c = ‘ ‘;
14         while (i >= 0 && j >= 0)
15         {
16             sum = (num1[i] - ‘0‘) + (num2[j] - ‘0‘) + mv_bit;
17             c = ‘0‘ + (sum % 10);
18             mv_bit = sum/10;
19             res_str = c + res_str;
20             --i;
21             --j;
22         }
23         if(i == j)//num1, num2位数相等,一起遍历完
24         {
25             return mv_bit == 0 ? res_str:‘1‘ + res_str;
26         }
27         //一个字符串比另一个 多出的字符
28         string front_str = i >= 0?num1.substr(0,i+1):num2.substr(0,j+1);
29         if(mv_bit == 0)
30         {
31             return front_str + res_str;
32         }
33         return addStrings(front_str,"1") + res_str; 
34     }
35 };