LC-415 字符串相加（大数相加）

目标：

输入两个十进制数的字符串，输出它们的和

思路：

这道题有两个思路，一个是最容易想到的，就是类小学加法，一位一位的加，记录是否要进位，一位一位地输出结果。

而我的思路是利用字符串分割，将比较大的数字加法，分成每8位的相加，这里就要处理好分割计算后，进位的处理，以及字符串是否需要补零的操作（因为有可能8位中前几位是0，使得加出来的数值再变回字符串时位数不足8）。

最终这种分割方法可以做到12ms通过测试。

代码：

 1 class Solution {
 2 public:
 3     string addStrings(string num1, string num2) {
 4         // make the longer number become num1
 5         if (num2.length() > num1.length()) {
 6             string temp = num1;
 7             num1 = num2;
 8             num2 = temp;
 9         }
10         string answer;
11         bool flag = false; // flag as a carry signal
12         // judge if it is a large num that over an int range
13         if (num2.length() > 8) {
14             // split the num in 8 bits each group then do the sum
15             for (int i = num2.length(), j = num1.length(); i > 0; i -= 8, j -= 8) {
16                 int len = 8;
17                 // if there are no more can split
18                 if (i < 8)
19                     len = i;
20                 string sub2 = num2.substr(i - len, len);
21                 string sub1 = num1.substr(j - len, len);
22                 int n2 = stoi(sub2);
23                 int n1 = stoi(sub1);
24                 int sum = n1 + n2;
25                 // check carry
26                 if (flag)
27                     sum++;
28                 string num = to_string(sum);
29                 // check if sum is over range, need a carry signal
30                 if (sum >= pow(10, len)) {
31                     flag = true;
32                     num.erase(0, 1);
33                 }
34                 else
35                     flag = false;
36                 // make complement of lack of 0
37                 num.insert(0, len - num.length(), '0');
38                 // append the number
39                 answer.insert(0, num);
40             }
41         }
42         else {
43             // no need to split
44             int len = num2.length();
45             string sub1 = num1.substr(num1.length() - len, len);
46             int n2 = stoi(num2);
47             int n1 = stoi(sub1);
48             int sum = n1 + n2;
49             string num = to_string(sum);
50             // check if sum is over range, need a carry signal
51             if (sum >= pow(10, len)) {
52                 flag = true;
53                 num.erase(0, 1);
54             }
55             answer = num;
56         }
57         // deal to the rest of longer number1
58         string prefix = num1.substr(0, num1.length() - num2.length());
59         if (flag) {
60             int i = prefix.length() - 1;
61             char new_digit = 0, digit;
62             while ((new_digit == '0' || new_digit == 0) && i >= 0) {
63                 digit = prefix[i];
64                 new_digit = ((digit - '0' + 1) % 10) + '0';
65                 prefix[i] = new_digit;
66                 i--;
67             }
68             if (i == -1 && (new_digit == '0' || new_digit == 0)) {
69                 prefix.insert(0, 1, '1');
70             }
71         }
72         return prefix + answer;
73     }
74 };