动态规划——392. 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.

判断s所有字母是否在t中先后不变的出现（不必连续）

解题思路

下标i、j表示s和t的当前字母，s［i］＝＝t［j］时i++；

最后判断i是否等于s.size（）

代码如下

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

题目简单，可是连一个状态转移方程都没有，很不像动态规划，但是，想了许久还没想到其他方法（以后想到再来更新）