摘要:題目鏈接只要里面當前的字符可以和里的字符匹配,的就很長,用字典存起來很大,用
Is Subsequence
題目鏈接:https://leetcode.com/problems...
greedy, 只要s里面當前的字符可以和t里的字符匹配,s的index就+1
public class Solution { public boolean isSubsequence(String s, String t) { // 2 points, greedy int i = 0, j = 0; while(i < s.length() && j < t.length()) { if(s.charAt(i) == t.charAt(j)) i++; j++; } return i == s.length(); } }
follow up: string很長,用字典存起來
dict很大,用trie
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