题目

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?

题意

关键概念：一个字符串的子序列是指，从最初的字符串中删除一些字符生成一个新的字符串，并且不改变剩下字符的相对顺序。

分析及解答

说明：

需要思考
1.为何这个问题适合使用贪心策略？
2.如何将子问题与原问题 通过贪心策略联系起来？

分析：

1. 【主要变量】问题中涉及的两个主要变量，一个源串，一个是目标串。
2. 【变量有序性】源串以及目标串都具有有序性，即我不能利用排序带来的优化。
3. 【问题分解+无后效性】分解的子问题之间应该具有一定的独立性（无后效性），两个指针标识的位置意味着将问题分解成了新的源串和目标串，后面的子问题中不需要考虑之前的子问题的内容，即后面的子问题不需看到之前的问题。

public boolean isSubsequence(String s, String t) {
		int i,j;
		for(i=0,j=0;i < s.length() && j < t.length();){
			if(s.charAt(i) == t.charAt(j)){
				i++;
				j++;
			}else{
				j++;
			}
		}
		if(i == s.length()){
			return true;
		}
		return false;
	}