宣告：原題目轉載自LeetCode，解答部分為原創

Problem ：

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"

Example 1:
s = "abc", t = "ahbgdc"

Return true.

Example 2:
s = "axc", t = "ahbgdc"

Return false.

Solution :

         思路：可以用動態規劃的思想來解決，定義bool函式L( i, j ) 返回子串s[ 0, i ] 是否為 t[ 0, j ] 的子序列。則L( i, j )的狀態轉換方程為

if  i = -1 : L( i, j ) = true;

else if   j = -1 : L( i, j ) = false;

else if  s[i] = t[j] : L( i , j ) = L( i -1, j - 1);

else : L( i , j ) = L( i, j - 1);

               程式碼如下：

#include<iostream>
using namespace std;

class Solution {
public:
/*    bool isSubsequence(string s, string t) {		
		return judge(s, t, s.size() - 1, t.size() - 1);
    }
    
    bool judge(string s, string t, int pos_s, int pos_t)
    {
    	if(pos_s == -1)
    		return true;
    	else if(pos_t == -1)
    		return false;
    	else if(s[pos_s] == t[pos_t])
    		return judge(s, t, pos_s - 1, pos_t - 1);
    	else
    		return judge(s, t, pos_s, pos_t - 1);
	}
*/	
	bool isSubsequence(string s, string t) {
    	int pos_s = 0;
    	for(int i = 0 ; i < t.size() && pos_s < s.size(); i ++)
    	{
            if(s[pos_s] == t[i])
            pos_s ++;
        }
        
        if(pos_s == s.size())
            return true;
        else
            return false;
    }
};


int main()
{
	Solution text;
	cout << text.isSubsequence("ace", "ahencgoe") << endl;
	return 0;
}