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...
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...
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 horizontally or vertically neighboring. The sam...
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 horizontally or vertically neighboring. The same letter cell may not be used more than once. For example, Given boar...
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){...
class Solution{public:boolexist(vector<vector<char>>&board,string word){for(inti=0;i<board.size();i++)for(intj=0;j<board[0].size();j++)if(board[i][j]==word[0])if(backtrack(board,word,i,j,board.size(),board[0].size(),0,word.size()))returntrue;returnfalse;}private:boolbac...
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
地址:https://oj.leetcode.com/problems/word-search/ 算法:用DFS加回溯法来解决。首先,从board的每一个位置开始搜索,假设当前遍历到第(i,j)个位置,如果word的第一个字符等于board[i][j],那么将以遍历的位置置为true,然后调用函数existCore进行DFS,existCore函数的最后三个参数表示,当前遍历到第(i,j)个位置...
class Solution: def __init__(self): self.root = None self.res = [] def findWords(self, board, words): """ :type board: List[List[str]] :type words: List[str] :rtype: List[str] """ # 如果为空则直接返回 if not board or not words: return [] ...
/** @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"...