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LeetCode 415 Add Strings

2024-08-18 06:46:40   218人阅读

Problem：

Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2.

  1. The length of both num1 and num2 is < 5100.
  2. Both num1 and num2 contains only digits 0-9.
  3. Both num1 and num2 does not contain any leading zero.
  4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

Summary:

以字符串的形式求两个整形数相加。

Analysis:

这道题是一道常规题，并不难，只需一位一位相加，注意在有进位的时候加上进位即可。

 1 class Solution {
 2 public:
 3     string addStrings(string num1, string num2) {
 4         int len1 = num1.size(), len2 = num2.size();
 5         int i = len1 - 1, j = len2 - 1, k = 0, carry = 0;
 6         char res[5200];
 7         
 8         while (i >= 0 && j >= 0) {
 9             int sum = num1[i] + num2[j] - 2 * ‘0‘ + carry;
10             if (sum > 9) {
11                 sum -= 10;
12                 carry = 1;
13             }
14             else {
15                 carry = 0;
16             }
17             
18             res[k++] = sum + ‘0‘;
19             i--; j--;
20         }
21         
22         while (i >= 0) {
23             int sum = num1[i] + carry - ‘0‘;
24             if (sum > 9) {
25                 sum -= 10;
26                 carry = 1;
27             }
28             else {
29                 carry = 0;
30             }
31             
32             res[k++] = sum + ‘0‘;
33             i--;
34         }
35         
36         while (j >= 0) {
37             int sum = num2[j] + carry - ‘0‘;
38             if (sum > 9) {
39                 sum -= 10;
40                 carry = 1;
41             }
42             else {
43                 carry = 0;
44             }
45             
46             res[k++] = sum + ‘0‘;
47             j--;
48         }
49         
50         if (carry) {
51             res[k++] = ‘1‘;
52         }
53         
54         res[k++] = ‘\0‘;
55         
56         for (i = 0; i < k / 2; i++) {
57             char tmp = res[i];
58             res[i] = res[k - i - 2];
59             res[k - i - 2] = tmp;
60         }
61         
62         string resStr = string(res);
63         
64         return resStr;
65     }
66 };

以上为我写的原始代码，代码过于冗长，下面是参考网上大牛的代码后优化的代码，代码简介了很多：

 1 class Solution {
 2 public:
 3     string addStrings(string num1, string num2) {
 4         int len1 = num1.size(), len2 = num2.size();
 5         int i = len1 - 1, j = len2 - 1, carry = 0;
 6         string res = "";
 7         
 8         while (i >= 0 || j >= 0) {
 9             int sum1 = i >= 0 ? num1[i--] - ‘0‘ : 0;
10             int sum2 = j >= 0 ? num2[j--] - ‘0‘ : 0;
11             
12             int sum = sum1 + sum2 + carry;
13             res.insert(res.begin(), sum % 10 + ‘0‘);
14             carry = sum / 10;
15         }
16         
17         return carry ? "1" + res : res;
18     }
19 };

 

LeetCode 415 Add Strings


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