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?
Credits: Special thanks to @pbrother for adding this problem and creating all test cases.
第一种方法，双指针直接遍历法，较之第二种方法慢。从前向后遍历t，遇到跟s指针下字符相等，s的指针向后移一位，当s的指针等于s的长度时，返回true。代码如下： public class Solution { public boolean isSubsequence(String s, String t) { if (s.length() == 0) { return true; } int indexS = 0, indexT = 0; for (char ch: t.toCharArray()) { if (ch == s.charAt(indexS)) { indexS ++; } if (indexS == s.length()) { return true; } } return false; } }第二种方法，利用的是string的indexOf函数。遍历的是较短的s，根据s的各个字符串去t里面找，如果找不到当前的字符，立马返回false，如果可以顺利遍历完s，返回true。代码如下： public class Solution { public boolean isSubsequence(String s, String t) { if (t.length() < s.length()) { return false; } int prev = 0; for (char ch: s.toCharArray()) { prev = t.indexOf(ch, prev); if (prev == -1) { return false; } prev ++; } return true; } }