classSolution{public:intlongestStrChain(vector<string>& words){intn = words.size(), res =1; sort(words.begin(), words.end(), [](string& a,string& b){returna.size() < b.size(); }); unordered_map<string,int> dp;for(stringword : words) { dp[word] =1;for(inti =0; i < w...
package leetcode import "sort" func longestStrChain(words []string) int { sort.Slice(words, func(i, j int) bool { return len(words[i]) < len(words[j]) }) poss, res := make([]int, 16+2), 0 for i, w := range words { if poss[len(w)] == 0 { poss[len(w)] = i ...
leetcode1048 Longest String Chain 1048 Longest String Chain Given a list of words, each word consists of English lowercase letters. Let's say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2. For example, "abc"...
Longest String Chain 查看原文 leetcode【3】最长不重复子串---【Python】【字典】 问题描述 给定一串字符串,输出最长不重复的子串输入输出 代码实现 笔试题 一,搜索连通域 二,简单二叉树的实现 三,找列表里的最长字符串四,找两个字符串的最长公共字符串五,如题 六,如题 七,如题 八,如题,采用贪心算法 九...
[leetcode] 1048. Longest String Chain Description Given a list of words, each word consists of English lowercase letters. Let’s say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2. For example, “abc” is a...
leetcode.com/problems/longest-string-chain/ 构造两个map,一个用来把长度相同的单词放在一起,一个用来统计到此为止的最长的长度 class Solution { public: int longestStrChain(vector<string>& words) { int N=words.size(); if(N<=1) return N; int max_length=0; // 记录最长单词的长度 for(...
1048. Longest String Chain PcPc Code for food 来自专栏 · Algorithms Barn Given a list of words, each word consists of English lowercase letters. Let's say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2. For...
Leetcode_5 Longest Palindromic Substring 一道难度为medium的题目,原题地址:https://leetcode.com/problems/longest-palindromic-substring/,用动态规划即可解决,但是效率不是特别高,执行时间为93ms。 题目: Given a string s, find the longest palindromic subs... ...
Leetcode 1048. Longest String Chain 编程算法 **解析:**Version 1,先根据字符串长度对数组排序,然后根据长度分到不同的组里,按长度遍历组,如果下一组的字符串长度比当前组多1个,则遍历两组的所有元素,满足条件前辈子串,则下一组子串的字符链长度在当前子串长度的基础上加1,其实就是一个广度优先搜索的过程。
File metadata and controls Code Blame 46 lines (39 loc) · 1.27 KB Raw class Solution: def lengthOfLongestSubstring(self, s): """ :type s: str :rtype: int """ if not s: return 0 # sliding window that keeps track of currentBest O(N^2) curSet = set() curInd = [...