classSolution {publicbooleanisSubsequence(String s, String t) {intlen1 =s.length();intlen2 =t.length();boolean[][] dp =newboolean[len1+1][len2+1];for(inti = 0; i < dp[0].length; i++) { dp[0][i] =true; }for(inti = 1; i < len1+1; i++) {for(intj = 1; j <...
leetcode 392. Is Subsequence 一、题意 给定字符串 s 和 t ,判断 s 是否为 t 的子序列。 进阶: 如果有大量输入的 S,称作 S1, S2, … , Sk 其中 k >= 10亿,你需要依次检查它们是否为 T 的子序列。在这种情况下,你会怎样改变代码? 二、解法 解法一: 双指针 子序列只要保证在t中,s中字符相对...
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+...
Is Subsequence判断子序列【Easy】【Python】【双指针】 Problem LeetCode 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 ...
Can you solve this real interview question? Is Subsequence - Given two strings s and t, return true if s is a subsequence of t, or false otherwise. A subsequence of a string is a new string that is formed from the original string by deleting some (can b
LeetCode "Is Subsequence" There are 3 possible approaches: DP, divide&conquer and greedy. And apparently, DP has O(n^2) complexity (TLE), DivideConquer can only be worse. Greedy can be O(n)! And its code is amazingly brief: classSolution {public:boolisSubsequence(strings,stringt) {...
392. Is Subsequence # 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
动态规划。设dp[i][j]为S以i结尾的子串是否是T以j结尾的子串的subsequence。首先corner case是当S为空的时候,dp[0][j] = true。其次开始扫描,如果匹配(s[i] == t[j]),则DP值跟之前一位的DP值相同(dp[i][j] == dp[i - 1][j - 1]);如果不匹配,则dp[i][j] = dp[i][j - 1]。
题目描述: LeetCode 392. Is Subsequence 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...
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...