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