题目描述：
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.
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完成两个字符串相乘。
思路：
1. 实现1个字符与字符串的相乘函数
2. 实现两个字符串相加的函数
3. 遍历num1[]，将num1[i]与num2相乘得到r，使用result累加保存r (r需要每次追加'0')。
实现代码：public class Solution {
public string Multiply(string num1, string num2)
{
if(string.IsNullOrEmpty(num1) || string.IsNullOrEmpty(num2)){
return "";
}
if(num1 == "0" || num2 == "0"){
return "0";
}
var p1 = num1.Length - 1;
var p2 = num2.Length - 1;
var result = string.Empty;
var pow = "";
for(var i = p1;i >= 0;i --){
var r = Multiply(num1[i], num2) + pow;
if(result == string.Empty){
result = r;
}
else{
result = Add(r, result);
}
pow += "0";
}
return result;
}
private string Add(string s1, string s2)
{
var p1 = s1.Length - 1;
var p2 = s2.Length - 1;
var carry = 0;
var result = "";
while(p1 >= 0 || p2 >= 0){
var n1 = 0;
var n2 = 0;
if(p1 >= 0){
n1 = s1[p1] - '0';
}
if(p2 >= 0){
n2 = s2[p2] - '0';
}
var r = n1 + n2 + carry;
carry = r / 10;
result = r % 10 + result;
p1 --;
p2 --;
}
if(carry > 0){
result = carry + result;
}
return result;
}
private string Multiply(char c, string s)
{
var d = c - '0';
if(d == 0){
return "0";
}
var carry = 0;
var result = "";
for(var i = s.Length - 1;i >= 0; i--){
var n = s[i] - '0';
var r = d * n + carry;
carry = r / 10;
result = r % 10 + result;
}
if(carry > 0){
result = carry + result;
}
return result;
}
}