LeetCode: Multiply Strings

最新推荐文章于 2025-09-05 17:57:56 发布

weixin_30709061

最新推荐文章于 2025-09-05 17:57:56 发布

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CC 4.0 BY-SA版权

文章标签： c#

原文链接：http://www.cnblogs.com/yingzhongwen/archive/2013/04/10/3011603.html

本文介绍了一种字符串乘法的算法实现，通过将字符串转换为整数进行相乘，并使用了C++和C#两种语言进行编码展示。该算法考虑了进位处理及最终结果的字符串格式化。

一点小失误，少数次过

 1 class Solution {
 2 public:
 3     string multiply(string num1, string num2) {
 4         vector<int> mul(num1.size() + num2.size());
 5         for (int i = 0; i < num1.size(); i++) {
 6             for (int j = 0; j < num2.size(); j++) {
 7                 mul[num1.size()-1-i+num2.size()-1-j] += (num1[i] - '0') * (num2[j] - '0');
 8             }
 9         }
10         int carry = 0;
11         for (int i = 0; i < mul.size(); i++) {
12             mul[i] += carry;
13             carry = mul[i] / 10;
14             mul[i] %= 10;
15         }
16         string ans;
17         for (int i = mul.size()-1; i >= 0; i--) {
18             if (mul[i] == 0) {
19                 if (ans.size() > 0) ans += ('0' + mul[i]);
20             }
21             else ans += ('0' + mul[i]);
22         }
23         return ans.size() == 0? "0" : ans;
24     }
25 };

 1 public class Solution {
 2     public string Multiply(string num1, string num2) {
 3         int[] mul = new int[num1.Length + num2.Length];
 4         for (int i = 0; i < num1.Length; i++) {
 5             for (int j = 0; j < num2.Length; j++) {
 6                 mul[num1.Length-1-i+num2.Length-1-j] += (num1[i] - '0') * (num2[j] - '0');
 7             }
 8         }
 9         int c = 0;
10         for (int i = 0; i < mul.Length; i++) {
11             mul[i] += c;
12             c = mul[i] / 10;
13             mul[i] %= 10;
14         }
15         string ans = "";
16         for (int i = mul.Length-1; i >= 0; i--) {
17             if (mul[i] == 0) {
18                 if (ans.Length > 0) ans += mul[i];
19             }
20             else ans += mul[i];
21         }
22         return ans.Length == 0? "0" : ans;
23     }
24 }