接下来怎么dp才是难点,对于message中的位置i, 如果这个位置上的数字是1到9的那么这个数字和dp[i-1]是一种组合,如果这个数字和前一个构成的数在10到26之间,那么这两个数字和dp[i-2]又能构成一种decode way。 importjava.util.*;publicclassLeetCode{publicstaticvoidmain(String[] args){ Scanner sc=newSca...
recuision最深是k层,recursive部分空间复杂度应该是O(k) + O(m*n)(visit array) View Code Solution2: 与解1是一样的,但我们可以省掉O(m*n)的空间复杂度。具体的作法是:在进入DFS后,把访问过的节点置为'#',访问完毕之后再置回来即可。 View Code December 16th, 2014 重写,简洁一点点,11分钟1次AC...
http://bookshadow.com/weblog/2015/05/19/leetcode-word-search-ii/ 这里的trie用child dict表示子节点,key是letter,value是node。我们要理解成每个node是没有value的,所以父节点到子节点的path才对应一个letter,即这里child dict的key。 借由下图,解释一下trie的delete操作。以前以为isWord只可能在叶子节点才会...
日期 题目地址:https://leetcode.com/problems/word-search/description/ 题目描述 Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where “adjacent” cells are those horizontall...
/** @lc app=leetcodeid=212 lang=cpp** [212] Word Search II** https://leetcode.com/problems/word-search-ii/description/** algorithms* Hard (29.77%)* Likes: 1405* Dislikes: 78* Total Accepted: 132.7K* Total Submissions: 440.4K* Testcase Example: '[["o","a","a","n"],["e"...
1. Description Word Search II 2. Solution class Solution{public:vector<string>findWords(vector<vector<char>>&board,vector<string>&words){vector<string>result;introws=board.size();intcolumns=board[0].size();unordered_set<string>s;for(string word:words){s.insert(word);}for(string word:s){...
参考LeetCode #208 Implement Trie (Prefix Tree) 实现 Trie (前缀树)和LeetCode #79 Word Search 单词搜索 将words数组中的所有 word插入到前缀树中 再在board数组中进行 dfs搜索 时间复杂度O(mn * 3 ^ l), 空间复杂度O(lk), 其中 m和 n分别是 board数组的长宽, l是 words数组中的单词长度, k为 wo...
思路:使用dfs. 先在矩阵中找到word0],若找到,就执行helper函数继续查找。递归结束条件是 index == word.length(); 如果是超过边界或者board[i != word.charAt(index) 就return false。 为防止same letter cell may be used more than once, 要标记已经访问过的letter为 #. 每次递归结束以后还要还原以防影响...
Can you solve this real interview question? Word Break II - Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order. Note
public class Solution { public ArrayList<ArrayList<String>> findLadders(String start, String end, HashSet<String> dict) { dict.add(end); ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>(); if (start == null || end == null || start.length() != end.length()) {...