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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.
给出两个字符串，返回对应数字想乘后的字符串，由于这个字符串可能很大，所以不能采用一般的乘法，这里用的方法是模拟手工的乘法运算，算法
本身很简单，就是当时写的时候有些很小的细节搞错了，找了很久的错。啊啊啊啊啊，要细心啊。大妈如下，没什么好说的：
1 class Solution { 2 public: 3 string multiply(string num1, string num2) { 4 if(num1 == "0" || num2 == "0") return "0"; 5 int steps = 0; 6 int pos = 0; 7 int flag = 0; 8 int val = 0; 9 string result = ""; 10 reverse(num1.begin(), num1.end()); 11 reverse(num2.begin(), num2.end()); 12 int len1 = num1.length(); 13 int len2 = num2.length(); 14 for(int i = 0; i < len1; ++i){ 15 pos = steps; 16 for(int j = 0; j < len2; ++j){ 17 val = (num1[i] - ‘0‘)*(num2[j] - ‘0‘) + flag; 18 if(result.size() <= pos){ 19 result.append(1, val%10 + ‘0‘); 20 }else{ 21 val += (result[pos] - ‘0‘); 22 result[pos] = val%10 + ‘0‘; 23 } 24 flag = val/10; 25 pos++; 26 } 27 if(flag > 0) 28 result.append(1, flag + ‘0‘); 29 flag = 0; 30 steps++; 31 } 32 reverse(result.begin(), result.end()); 33 return result; 34 } 35 };
LeetCode OJ：Multiply Strings (字符串乘法)
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4870089.html
