题目

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 board =

[
["ABCE"],
["SFCS"],
["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,

word = "ABCB", -> returns false.

题解

这道题分析看,就是一个词,在一行出现也是true,一列出现也是true,一行往下拐弯也是true,一行往上拐弯也是true,一列往左拐弯也是true,一列往右拐弯也是true。所以是要考虑到所有可能性,基本思路是使用DFS来对一个起点字母上下左右搜索,看是不是含有给定的Word。还要维护一个visited数组,表示从当前这个元素是否已经被访问过了,过了这一轮visited要回false,因为对于下一个元素,当前这个元素也应该是可以被访问的。

代码如下:

 1     public boolean exist(char[][] board, String word) {
 2         int m = board.length;  
 3         int n = board[0].length;  
 4         boolean[][] visited = new boolean[m][n];  
 5         for (int i = 0; i < m; i++) {  
 6             for (int j = 0; j < n; j++) {  
 7                 if (dfs(board, word, 0, i, j, visited))  
 8                     return true;  
 9             }  
         }  
         return false;  
     }
     
     public boolean dfs(char[][] board, String word, int index, int rowindex, int colindex, boolean[][] visited) {  
         if (index == word.length())  
             return true;  
         if (rowindex < 0 || colindex < 0 || rowindex >=board.length || colindex >= board[0].length)  
             return false;  
         if (visited[rowindex][colindex])  
             return false;  
         if (board[rowindex][colindex] != word.charAt(index))  
             return false;  
         visited[rowindex][colindex] = true;  
         boolean res = dfs(board, word, index + 1, rowindex - 1, colindex,  
                 visited)  
                 || dfs(board, word, index + 1, rowindex + 1, colindex, visited)  
                 || dfs(board, word, index + 1, rowindex, colindex + 1, visited)  
                 || dfs(board, word, index + 1, rowindex, colindex - 1, visited);  
         visited[rowindex][colindex] = false;  
         return res;  
             }

Reference:http://blog.csdn.net/yiding_he/article/details/18893621

Word Search leetcode java的更多相关文章

  1. Word Search [LeetCode]

    Problem Description: http://oj.leetcode.com/problems/word-search/ Basic idea: recursively go forward ...

  2. Word Ladder leetcode java

    题目: Given two words (start and end), and a dictionary, find the length of shortest transformation se ...

  3. Word Break leetcode java

    题目: Given a string s and a dictionary of words dict, determine if s can be segmented into a space-se ...

  4. Java for LeetCode 212 Word Search II

    Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...

  5. LeetCode解题报告—— Word Search & Subsets II & Decode Ways

    1. Word Search Given a 2D board and a word, find if the word exists in the grid. The word can be con ...

  6. [LeetCode] 79. Word Search 单词搜索

    Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from l ...

  7. [LeetCode] 212. Word Search II 词语搜索 II

    Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...

  8. [LeetCode] Word Search II 词语搜索之二

    Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...

  9. [LeetCode] Word Search 词语搜索

    Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from l ...

随机推荐

  1. 解决org.apache.jasper.JasperException: org.apache.jasper.JasperException: XML parsing error on file org.apache.tomcat.util.scan.MergedWebXml

    1.解决办法整个项目建立时采用utf-8编码,包括代码.jsp.配置文件 2.并用最新的tomcat7.0.75 相关链接: http://ask.csdn.net/questions/223650

  2. 7.30 正睿暑期集训营 A班训练赛

    目录 2018.7.30 正睿暑期集训营 A班训练赛 T1 A.蔡老板分果子(Hash) T2 B.蔡老板送外卖(并查集 最小生成树) T3 C.蔡老板学数学(DP NTT) 考试代码 T2 T3 2 ...

  3. Atcoder Tenka1 Programmer Contest 2019 题解

    link 题面真简洁 qaq C Stones 最终一定是连续一段 . 加上连续一段 # .直接枚举断点记录前缀和统计即可. #include<bits/stdc++.h> #define ...

  4. Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) E. Goods transportation 动态规划

    E. Goods transportation 题目连接: http://codeforces.com/contest/724/problem/E Description There are n ci ...

  5. Spring_Spring@Transactional

    Spring事务的传播行为 在service类前加上@Transactional,声明这个service所有方法需要事务管理.每一个业务方法开始时都会打开一个事务. Spring默认情况下会对运行期例 ...

  6. FireDAC 下的 Sqlite [9] - 关于排序

    SQLite 内部是按二进制排序, 可以支持 ANSI; FrieDAC 通过 TFDSQLiteCollation 支持了 Unicode 排序, 并可通过其 OnCompare 事件自定义排序. ...

  7. 独家专访|浙江执御:为何接受富安娜入股而不选VC?_深圳市跨境电子商务协会_新浪博客

    独家专访|浙江执御:为何接受富安娜入股而不选VC?_深圳市跨境电子商务协会_新浪博客   http://blog.sina.com.cn/s/blog_13cb5d69e0102vuvk.html

  8. [多问几个为什么]为什么匿名内部类中引用的局部变量和参数需要final而成员字段不用?(转)

    昨天有一个比较爱思考的同事和我提起一个问题:为什么匿名内部类使用的局部变量和参数需要final修饰,而外部类的成员变量则不用?对这个问题我一直作为默认的语法了,木有仔细想过为什么(在分析完后有点印象在 ...

  9. Unity中Web.Config文件的配置与调用

    在上一篇文章“Unit简单依赖注入”我们可以实现构造对象和被依赖对象之间的 松耦合,使我们的抽象层(Player)能够保持稳定,但是在并没有把客户类和Player类之间彻底解耦,即当我们不想使用MP3 ...

  10. USBDM Coldfire V2,3,4/DSC/Kinetis Debugger and Programmer -- MC9S08JS16

    Introduction The attached files provide a port of a combined TBLCF/DSC code to a MC9S08JS16 processo ...