43. Multiply Strings

Given two non-negative integers 

num1

 and 

num2

 represented
as strings, return the product of 

num1

 and 

num2

.

Note:
The length of both 

num1

 and 

num2

 is
< 110.
Both 

num1

 and 

num2

 contains
only digits 

0-9

.
Both 

num1

 and 

num2

 does
not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.
模拟乘法运算过程。字符串数字num1和num2没有前导零，处理方便一些。程序如下：

class Solution {
public StringBuilder multiplySingleChar(String num1, char ch){
int carry = 0， val = 0;
StringBuilder s = new StringBuilder();
for (int i = num1.length() - 1; i >= 0; -- i){
s.insert(0, ((num1.charAt(i) - '0')*(ch - '0') + carry)%10);
carry = ((num1.charAt(i) - '0')*(ch - '0') + carry)/10;
}
if (carry != 0){
s.insert(0, carry);
}
return s;
}

public String addString(String num1, StringBuilder num2){
int carry = 0;
int index1 = num1.length() - 1, index2 = num2.length() - 1;
StringBuilder sum = new StringBuilder();
while (index1 >= 0&&index2 >= 0){
sum.insert(0, (num1.charAt(index1) - '0' + num2.charAt(index2) - '0' + carry)%10);
carry = (num1.charAt(index1) - '0' + num2.charAt(index2) - '0' + carry)/10;
index1 --;
index2 --;
}
while (index1 >= 0){
sum.insert(0, (num1.charAt(index1) - '0' + carry)%10);
carry = (num1.charAt(index1) - '0' + carry)/10;
index1 --;
}
while (index2 >= 0){
sum.insert(0, (num2.charAt(index2) - '0' + carry)%10);
carry = (num2.charAt(index2) - '0' + carry)/10;
index2 --;
}
if (carry > 0){
sum.insert(0, carry);
}
return new String(sum);
}

public String multiply(String num1, String num2) {
if (num1.equals("0")||num2.equals("0")){
return "0";
}
String val = new String("0");
for (int j = num2.length() - 1; j >= 0; -- j){
StringBuilder tmp = multiplySingleChar(num1, num2.charAt(j));
for (int i = num2.length() - 1; i > j; -- i){
tmp.append('0');
}
val = addString(val, tmp);
}
return val;
}
}

本程序涉及大量的StringBuilder插入操作，效率较低，可通过字符数组的方式重新改写相关操作，效率会提高很多。