Leetcode-43. Multiply Strings

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

——————————————————————————————
Given two numbers represented as strings, return multiplication of the numbers as a string.

Note:

The numbers can be arbitrarily large and are non-negative.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.

这个题目最简单的就是用O(n^2)的时间效率，用代码实现我们手动这么算的。效果还不错Your runtime beats 40.75% of java submissions.

public class Solution {
public String multiply(String num1, String num2) {

if(num1.equals("0") || num2.equals("0")) return "0";
if(num1.equals("1")) return num2;
if(num2.equals("1")) return num1;
int[] results = new int[num1.length() + num2.length()];
for( int i = num1.length() - 1; i >= 0 ; i--)
for(int j = num2.length()-1 ;j >= 0 ; j --){
int char2num1 = num1.charAt(i) - '0';
int char2num2 = num2.charAt(j) - '0';

int carryNumber = char2num1 * char2num2 + results[ i + j + 1];

results[i + j + 1] = carryNumber % 10;
results[i + j ] = results[i + j] + carryNumber / 10;
}

StringBuffer sb = new StringBuffer("");
for(int i = 0 ; i< results.length ; i ++) {
if (i == 0 && results[i] == 0) continue;
sb.append(results[i]);
}
return sb.toString();
}
}