LeetCode#43. Multiply Strings

题目：

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2.

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

题意解析：

高精度乘法，就是通过像小学那样列数式相乘，将每一层得到的数式相加，并获得最终的结果。

例如333乘55

3 3 3

   5 5

      1 6 6 5

   1 6 6 5

   1 8 3 1 5

通过将每一层得到的数式存起来，再相加就可以得到最终的结果，空余的地方以0作为补充。

一种c++的代码实现如下：

#include<iostream>#include<string>using namespace std;class Solution {public:    string multiply(string num1, string num2) {    //若有一个数为0，则立即返回0     if(num1 == "0" || num2 == "0") return "0";        int n1 = num1.length();        int n2 = num2.length();string *process = new string[n2];//结果，和反向结果，为了表示方便，用反向结果进行运算，将反向结果反向已得到最终的结果 string result, invert_result;//将num2当做被乘数，则num2有多少位就会得到多少个数式，将其分别保存下来，并做初始化 for(int i = 0; i < n2; i++) {// 每个数式的长度的定义，由两个乘数的长度加1确定 for(int j = 0; j < n1+n2+1; j++) {process[i] += '0';}}//乘法运算 for(int i = 0; i < n2; i++) {int k = i;//进位 int carry = 0;for(int j = n1-1; j >= 0; j--) {//乘数 process[i][k++] = ( ( (num2[n2-1-i]-'0')*(num1[j] - '0') ) % 10 + carry ) % 10 + '0';//获得进位 carry = ( (num2[n2-1-i]-'0')* (num1[j] - '0') ) / 10 + ( ( (num2[n2-1-i]-'0')*(num1[j] - '0') ) % 10 + carry )/10;}//最开头的进位 if(carry != 0) process[i][k] = carry+'0';}int min_num_length = process[n2-1].length();/*for(int i = 0; i < n2; i++) {cout<<process[i]<<endl;}*///将所有得到的舒适相加，获得最终的结果 if(n2 != 1) {for(int j = 0; j < n1+n2; j++) invert_result += '0';int carry = 0;int last_carry = 0;for(int i = 0; i < min_num_length; i++) {carry = 0;for(int k = 0; k < n2-1; k++) {if(k == 0) {invert_result[i] = ( (process[k][i]-'0')+(process[k+1][i]-'0') )%10+'0';carry += ( (process[k][i]-'0')+(process[k+1][i]-'0') ) / 10;} else {int old_num = (invert_result[i]-'0');invert_result[i] = ( old_num+(process[k+1][i]-'0') )%10+'0';carry += ( old_num+(process[k+1][i]-'0') ) / 10;}}if(last_carry != 0) {int old_num = (invert_result[i]-'0');invert_result[i] = ( (invert_result[i]-'0') + last_carry ) % 10+'0';carry += ( old_num+ last_carry ) / 10;}last_carry = carry;//cout<<carry<<" "<<last_carry<<endl;}} else{invert_result += process[0];}int start = invert_result.length()-1;//反向，并将开头的零去掉 while(invert_result[start] == '0') {start--;}for(int i = start; i >= 0; i--) {result+= invert_result[i];}//cout<<result<<endl;return result;     }};