[LeetCode] Multiply Strings

[Problem]
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.

[Analysis]

大数乘法，不要忘了最后一个进位，注意删除前导0

[Solution]

class Solution {
public:
// add num2 to num1, make sure num1 has enough space
string reverseAdd(string num1, string num2){
int len1 = num1.length();
int len2 = num2.length();
int carry = 0, i;
for(i = 0; i < len1; ++i){
int sum = 0;
if(i < len2){
sum = (num1[i]-'0') + (num2[i]-'0') + carry;
}
else{
sum = (num1[i]-'0') + carry;
}
carry = sum / 10;
sum = sum % 10;
num1[i] = char(sum + '0');
}
if(carry > 0){
num1[i] = char(carry + '0');
}
return num1;
}

// reverse and remove heading '0'
string reverseAndRemove(string num){
int len = num.length();
int i = len - 1;
string res = "";

// remove heading '0'
while(i >= 0 && num[i] == '0'){
i--;
}

// all of them are '0's
if(i < 0){
res.append("0");
}
else{
for(int j = i; j >= 0; --j){
res.append(num.substr(j, 1));
}
}
return res;
}

// multiply
string multiply(string num1, string num2) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
string res = "", tmp = "";
int len1 = num1.length();
int len2 = num2.length();

// initial
for(int i = 0; i < len1+len2; ++i){
res.append("0");
tmp.append("0");
}

// multiply
int i, j;
for(i = len2-1; i >= 0; --i){
// product of num1*num2[i]
string tmpProduct = tmp;

// process of num1*num2[i]
int carry = 0;
for(j = len1-1; j >= 0; --j){
int product = (num2[i]-'0') * (num1[j]-'0') + carry;
carry = product / 10;
product = product % 10;

// reverse set
tmpProduct[(len2-1-i)+(len1-1-j)] = char(product + '0');
}

// be careful the remained carry
if(carry > 0){
tmpProduct[(len2-1-i)+(len1-1-j)] = char(carry + '0');
}

// add temp product to res
// MIND the res and tmpProduct are reversed
res = reverseAdd(res, tmpProduct);
}

// reverse resul and remove heading '0'
return reverseAndRemove(res);
}
};

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