LeetCode 43. Multiply Strings

原题链接在这里：https://leetcode.com/problems/multiply-strings/

题目：

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"
Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

题解：

Method 1:通过这道题, 学到了BigInteger的用法, 他的constructor 可以直接从string建立BigInteger, 但要注意它的乘法API是bi1.multiply(bi2).

若是中间需要考虑溢出，还有bi1.intValue(), bi1.longValue()等API.

Method 2:若num1.length() = m, num2.length() = n, 那么结果的长度为m+n-1(没有进位)，或者是m+n(有进位)。

以289*758为例:

生成数组d来储存成绩和结果，例子中的14，66，119，109，72行，然后取余进位.

为了方便起见，把原string reverse.

Note: corner case "0" * "1234", if either string starts with 0, return "0". Otherwise it would return "0000".

Time Complexity: O(len1*len2).

Space: O(len1+len2).

AC Java:

 1 import java.math.*;
 2 public class Solution {
 3     public String multiply(String num1, String num2) {
 4         /*Method 1
 5         if(num1 == null || num2 == null || num1.length() ==0 || num2.length() == 0)
 6             return "";
 7         BigInteger bi1 = new BigInteger(num1);
 8         BigInteger bi2 = new BigInteger(num2);
 9         
10         BigInteger res = bi1.multiply(bi2);
11         return res.toString();
12         */
13         
14         //Method 2
15         if(num1 == null || num1.length() == 0 || num2 == null || num2.length() == 0){
16             return "";
17         }
18         if(num1.charAt(0) == '0' || num2.charAt(0) == '0'){
19             return "0";
20         }
21         num1 = new StringBuilder(num1).reverse().toString();
22         num2 = new StringBuilder(num2).reverse().toString();
23         int len1 = num1.length();
24         int len2 = num2.length();
25         
26         int [] d = new int[len1 + len2];
27         for(int i = 0; i<len1; i++){
28             int a = num1.charAt(i) - '0';
29             for(int j = 0; j<len2; j++){
30                 d[i+j] += a * (num2.charAt(j)-'0');
31             }
32         }
33         
34         int cur = 0;
35         int carry = 0;
36         StringBuilder sb = new StringBuilder();
37         for(int i = 0; i<d.length; i++){
38             cur = (d[i]+carry)%10;
39             carry = (d[i]+carry)/10;
40             sb.insert(0,cur);
41         }
42         if(sb.charAt(0) == '0'){
43             sb.deleteCharAt(0);
44         }
45         return sb.toString();
46     }
47 }

类似Add Strings.