LeetCode 0392. Is Subsequence判断子序列【Easy】【Python】【双指针】 Problem LeetCode Given a stringsand a stringt, check ifsis subsequence oft. You may assume that there is only lower case English letters in bothsandt.tis potentially a very long (length ~= 500,000) string, andsis a sho...
classSolution(object):defisSubsequence(self,s,t):""":type s: str:type t: str:rtype: bool"""ifnotsandnott:returnTrueelifnotsandt:returnTrueelifnottands:returnFalsem,n=len(t),len(s)dp=[[Falseforjinrange(m+1)]foriinrange(2)]foriinrange(m+1):dp[0][i]=Trueforiinxrange(1,n+...
英文网址:392. Is Subsequence。 中文网址:392. 判断子序列。 思路分析 求解关键:这道题的解法其实蕴含了贪心算法的思想。 参考解答 参考解答1: public class Solution { public boolean isSubsequence(String s, String t) { int slen = s.length(); int tlen = t.length(); int sl = 0; int tl = ...
https://leetcode.com/contest/3/problems/is-subsequence/ 题目: s and a string t, check if s is subsequence of t. s and t. t is potentially a very long (length ~= 500,000) string, and s "ace"is a subsequence of"abcde"while"aec"Example 1: s ="abc", t ="ahbgdc"true.Example...
Python3 实现 常规双指针解法 # @author:leacoder # @des: 双指针解法 判断子序列 class Solution: def isSubsequence(self, s: str, t: str) -> bool: # 判断 s 是否为 t 的子序列。 # 指针 point_s 指向 s 第一个字符,指针 point_t 指向 t 第一个字符。逐一判断 point_s 所指向的字符是否在...
Leetcode - Is Subsequence publicclassSolution{publicbooleanisSubsequence(Strings,Stringt){if(s==null||t==null){returnfalse;}elseif(t.length()<s.length()){returnfalse;}intpre=0;for(inti=0;i<s.length();i++){charcurr=s.charAt(i);intindex=t.indexOf(""+curr,pre);if(index==-1){...
链接:https://leetcode-cn.com/problems/is-subsequence 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 思路 依次检查 s 中的每个字符是否出现在 t 中,并且要求 s 中后面的字符的在 t 中出现的位置对应递增。 当有很多个 s ,只有一 t 个时,可以考虑用字典对 t 建立索引。键为 t...
leetcode392. Is Subsequence 题目要求 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)....
classSolution {public:boolisSubsequence(strings,stringt) {if(s.empty())returntrue;inti =0, j =0;while(i < s.size() && j <t.size()) {if(s[i] == t[j]) ++i;++j; }returni ==s.size(); } }; 本文转自博客园Grandyang的博客,原文链接:是子序列[LeetCode] Is Subsequence,如需转...
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