动态规划--------Is Subsequence

题目描述：

Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:
s = "abc", t = "ahbgdc"

Return true.

Example 2:
s = "axc", t = "ahbgdc"

Return false.

Follow up:
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence.

In this scenario, how would you change your code?

import java.lang.*;
public class Solution {
    public boolean isSubsequence(String s, String t) {
      //此题应该挺简单的，但就是不懂哪里用到动态规划
     boolean result=false;
     for(int i=0;i<s.length();i++)
       { for(int j=i;j<t.length();j++)
         if(s.charAt(i)=t.charAt(j))
         break;
         if(i<s.length()&&j==t.length())
         return result;
         if(i==s.length())
         result=true;
       }
         
     return result;
        
    }
}

编译没有通过，但是暂时不知道错在哪里，，难道是charAt（）函数需要单独定义，不能直接调用。

换C++做法的

class Solution {
public:
    bool isSubsequence(string s, string t) {
        if (s.empty()) return true;
        int i = 0, j = 0;
        while (i < s.size() && j < t.size()) {
            if (s[i] == t[j]) {
                ++i; ++j;
            } else {
                ++j;
            }
        }
        return i == s.size();
    }
};
编译通过