LeetCode OJ：Multiply Strings (字符串乘法)

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。代码如下，没什么好说的：

 class Solution {
 public:
     string multiply(string num1, string num2) {
         if(num1 == "" || num2 == "") return "";
         int steps = ;
         int pos = ;
         int flag = ;
         int val = ;
         string result = "";
         reverse(num1.begin(), num1.end());
         reverse(num2.begin(), num2.end());
         int len1 = num1.length();
         int len2 = num2.length();
         for(int i = ; i < len1; ++i){
               pos = steps;
             for(int j = ; j < len2; ++j){
                 val = (num1[i] - '')*(num2[j] - '') + flag;
                 if(result.size() <= pos){
                     result.append(, val% + '');
                 }else{
                     val += (result[pos] - '');
                     result[pos] = val% + '';
                 }
                 flag = val/;
                 pos++;
             }
             if(flag > )
                 result.append(, flag + '');
             flag = ;
             steps++;
         }
         reverse(result.begin(), result.end());
         return result;
     }
 };

java的api真是恶心，处理字符串处理了半天，最后debug通过，往上一贴竟然是TLE的代码，代码在下面，应该是我对字符串的处理比较耗时间了，其实最好的应该是先讲String转换成List，List里面相对的API的函数多一点，String里面连最普通的reverse函数都没有。先马一下这种做法，有时间了再来写。下面的这个道理上应该没有任何问题，就是字符串处理的太慢了。

 public class Solution {
     public String multiply(String num1, String num2) {
         if(num1.equals("0") || num2.equals("0"))
             return (new String("0"));
         int step = 0;
         int pos = 0;
         int carry = 0;
         int val = 0;
         String result = new String("");
         num1 = ReverseStr(num1);
         num2 = ReverseStr(num2);
         for(int i = 0; i < num1.length(); ++i){ //java的api好乱啊,一会是length一会有是size()
             pos = step;
             for(int j = 0; j < num2.length(); ++j){
                 val = (num1.charAt(i)-'0')*(num2.charAt(j)-'0') + carry;
                 if(pos >= result.length())
                     result += (char)(val%10 + '0');
                 else{
                     val += (result.charAt(pos)-'0');
                     result = (new StringBuffer(result)).replace(pos, pos + 1, "" + (char)(val%10+'0')).toString();
                 }
                 carry = val/10;
                 pos++;
             }
             if(carry != 0)
                 result += (char)(carry +'0');
             carry = 0;
             step++;
         }
         return ReverseStr(result);

     }

     public String ReverseStr(String str){
         return (new StringBuffer(str)).reverse().toString();
     }

 }