题目标签：Greedy
设两个 pointers，s_index 和 t_index；
如果 s_index 和 t_index 位置上的字母一样，那么继续移动两个 pointers；
如果字母不一样，只移动 t_index；
具体看code。
Java Solution:
Runtime: 7 ms, faster than 80.12%
Memory Usage: 44.5 MB, less than 100.00%
完成日期：02/19/2020
关键点：two pointers
class Solution { public boolean isSubsequence(String s, String t) { if(s.length() == 0) return true; int s_index =0, t_index = 0; // use two pointers to find char in s and t while(t_index < t.length()) { if(s.charAt(s_index) == t.charAt(t_index)) { // if found the match, move s_index to next one s_index++; if(s_index == s.length()) return true; } t_index++; } return false; } }
参考资料：LeetCode Discuss
LeetCode 题目列表 - LeetCode Questions List
题目来源：https://leetcode.com/
LeetCode 392. Is Subsequence (判断子序列)
原文：https://www.cnblogs.com/jimmycheng/p/12334484.html