IT公司面试题 用网上的dp代码 挂了 请问正确的dp该怎么写

Question

面试官给我出了道老题， 我用了一亩三分地上的dp解答，来源如下
http://www.1point3acres.com/bbs/thread-145290-1-1.html

题目如下：
String s1 = "waeginsapnaabangpisebbasepgnccccapisdnfngaabndlrjngeuiogbbegbuoecccc";
String s2 = "a+b+c-";

s2的形式是一个字母加上一个符号，正号代表有两个前面的字符，负号代表有四个，也就是说s2其实是"aabbcccc"，不考虑invalid。
在s1中，找出连续或者不连续的s2，也就是说从s1中找出"aa....bb.....cccc"，abc顺序不能变，但是之间可以有零个或多个字符，返回共有多少个。在上面这个例子中，有四个。
结果测试sln.findMatches("aaaaaa", "a+a-") ，出来结果为0，不对，挂了
另外System.out.println(sln.findMatches("waeginsapnaabangpisebbasccepgnccccapisdnfngaabndlrjngeuiogbbegbuoecccc", "a+b+c+c-")); 跑出来是5，感觉正确结果应该是2
不知道用dp正确的解答应该是什么？有人能给出正确的dp代码吗？

以下是从拷贝的测试用例不通过的代码：

 public class Solution {
        public int findMatches(String s1, String s2){
       int len1 = s1.length(), len2 = s2.length();
       if (len2 > len1) return 0;
       if (len2 == 0 || len2 %2 != 0) return 0;
       // DP matches each pattern
       // number of matches between s1.substring(0, i + 1) and s2.substring(j * 2, j * 2 + 2)
       int[][] dp = new int[len1 + 1][len2 / 2 + 1];
       // no match for dp[0][j]
       for (int i = 1; i < len1; i++){
           dp[i + 1][0] = 1;
           for (int j = 0; j < len2 / 2; j++){
               dp[i + 1][j + 1] = dp[i][j + 1];
               if (isMatch(s1, s2, i, j)){
                   if (s2.charAt(2*j + 1) == '+')
                       dp[i + 1][j + 1] += dp[i - 1][j];
                   else
                       dp[i + 1][j + 1] += dp[i - 3][j];
               }
           }
       }
       return dp[len1][len2 / 2];
  }

   boolean isMatch(String s1, String s2, int i, int j){
       char c = s2.charAt(j * 2);
       char p = s2.charAt(j * 2 + 1);
       int len = p == '+' ? 2 : 4;
       if (i - len < -1) return false;
       for (int h = i - len + 1; h <= i; h++){
           if (s1.charAt(h) != c) return false;
       }
       return true;
   }

   public static void main(String[] args){
           Solution sln = new Solution();

           System.out.println(sln.findMatches("waeginsapnaabangpisebbasepgnccccapisdnfngaabndlrjngeuiogbbegbuoecccc", "a+b+c-"));  // 4

           System.out.println(sln.findMatches("waeginsapnaabangpisebbasccepgnccccapisdnfngaabndlrjngeuiogbbegbuoecccc", "a+b+c+c-")); // 5 ？？？  I think it should be 2
           System.out.println(sln.findMatches("aaaaaa", "a+a-"));//0  wrong!
   }
}

Answer 1

拿C#给你写了一个

 using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading.Tasks;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            string s1 = "waeginsapnaabangpisebbasepgnccccapisdnfngaabndlrjngeuiogbbegbuoecccc";
            string s2 = "a+b+c-";
            List<string> list = new List<string>();
            for (int i = 0; i < s2.Length; i += 2)
            {
                list.Add(new string(s2[i], s2[i + 1] == '+' ? 2 : 4));
            }
            List<List<int>> idx = list.Select(x => new List<int>()).ToList();
            for (int i = 0; i < s1.Length; i++)
            {
                for (int j = 0; j < list.Count; j++)
                {
                    if (i <= s1.Length - list[j].Length && s1.Substring(i, list[j].Length) == list[j])
                    {
                        idx[j].Add(i);
                    }
                }
            }
            List<List<int>> result = idx[0].Select(x => new List<int>() { x }).ToList();
            for (int i = 1; i < idx.Count; i++)
            {
                result = result.SelectMany(x => idx[i].Where(y => y >= x.Last() + list[i - 1].Length).Select(y => x.Concat(new int[] { y }).ToList())).ToList();
            }
            foreach (var item in result) Console.WriteLine(string.Join(",", item));
            Console.WriteLine(result.Count);
        }
    }
}