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:

import java.math.*;
public class Solution {
    public String multiply(String num1, String num2) {
        /*Method 1
        if(num1 == null || num2 == null || num1.length() ==0 || num2.length() == 0)
            return "";
        BigInteger bi1 = new BigInteger(num1);
        BigInteger bi2 = new BigInteger(num2);
        
        BigInteger res = bi1.multiply(bi2);
        return res.toString();
        */
        
        //Method 2
        if(num1 == null || num1.length() == 0 || num2 == null || num2.length() == 0){
            return "";
        }
        if(num1.charAt(0) == '0' || num2.charAt(0) == '0'){
            return "0";
        }
        num1 = new StringBuilder(num1).reverse().toString();
        num2 = new StringBuilder(num2).reverse().toString();
        int len1 = num1.length();
        int len2 = num2.length();
        
        int [] d = new int[len1 + len2];
        for(int i = 0; i<len1; i++){
            int a = num1.charAt(i) - '0';
            for(int j = 0; j<len2; j++){
                d[i+j] += a * (num2.charAt(j)-'0');
            }
        }
        
        int cur = 0;
        int carry = 0;
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i<d.length; i++){
            cur = (d[i]+carry)%10;
            carry = (d[i]+carry)/10;
            sb.insert(0,cur);
        }
        if(sb.charAt(0) == '0'){
            sb.deleteCharAt(0);
        }
        return sb.toString();
    }
}

类似Add Strings.