Sum—LeetCode-43 Multiply Strings

题目描述：

Given two non-negative integers num1 andnum2 represented as strings, return the product ofnum1
and num2.
Note:
The length of both num1 andnum2 is < 110.
Both num1 andnum2 contains only digits
0-9.
Both num1 andnum2 does not contain any leading zero.
You must not use any built-in BigInteger library orconvert the inputs to integer directly.

思想：基本的乘法思想，见代码：

public String multiply(String num1, String num2) {
char[] chars1 = num1.toCharArray();
char[] chars2 = num2.toCharArray();
char[][] data = new char[num2.length()][num1.length() + num2.length()];

for(int m = 0; m < num2.length(); ++m){
for(int n = 0; n < num1.length() + num2.length(); ++n){
data[m]
= '0';
}
}

int back = 0;
for(int i = chars2.length - 1; i >= 0; --i){
int k = num1.length() + num2.length() - 1;
int b = Integer.valueOf(String.valueOf(num2.charAt(i)));
int c = 0;
for(int j = chars1.length - 1; j >= 0; --j){
int a = Integer.valueOf(String.valueOf(num1.charAt(j)));
int sum = a * b + c;
c = sum / 10;
data[i][k - back] = String.valueOf(sum % 10).charAt(0);
--k;
}
if(c != 0){
data[i][k - back] = String.valueOf(c).charAt(0);
}
++back;
}
return addArray(data);
}

private String addArray(char[][] data) {
int c = 0;
int row = data.length, col = data[0].length;
int k = col;
char resChar[] = new char[col + 1];
for(int p = 0; p < col + 1; ++p) {
resChar[p] = '0';
}
for(int i = col - 1; i >= 0; --i) {
int sum = 0;
for(int j = 0; j < row; ++j) {
sum += Integer.valueOf(String.valueOf(data[j][i]));
}
sum += c;
resChar[k] = String.valueOf(sum % 10).charAt(0);
c = sum / 10;
--k;
}

int startIndex = 0, flag = 0;
for(int m = 0; m < resChar.length; ++m){
if(resChar[m] != '0'){
startIndex = m;
flag = 1;
break;
}
}
if(c != 0) {
resChar[startIndex - 1] = String.valueOf(c).charAt(0);
--startIndex;
}
if(flag == 0){
return "0";
}
return String.valueOf(resChar).substring(startIndex);
}