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LintCode-Word Search II

时间:2014-12-25 06:35:39      阅读:169      评论:0      收藏:0      [点我收藏+]

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Given a matrix of lower alphabets and a dictionary. Find all words in the dictionary that can be found in the matrix. A word can start from any position in the matrix and go left/right/up/down to the adjacent position. 

 

Example

Given matrix:

doaf
agai
dcan

and dictionary:

{"dog", "dad", "dgdg", "can", "again"}
 
return {"dog", "dad", "can", "again"}
 
Analysis:
DFS. For every word in the list, check every position of board, if the char at some position matches the first char in the word, then start a DFS at this position.
 
Solution:
 1 public class Solution {
 2     /**
 3      * @param board: A list of lists of character
 4      * @param words: A list of string
 5      * @return: A list of string
 6      */
 7     public ArrayList<String> wordSearchII(char[][] board, ArrayList<String> words) {
 8         ArrayList<String> res = new ArrayList<String>();
 9         if (board.length==0) return res;
10         int rowNum = board.length;
11         if (board[0].length==0) return res;
12         int colNum = board[0].length;
13 
14         for (int i=0;i<words.size();i++){
15             String word = words.get(i);
16             if (word.length()==0) continue;
17             for (int j=0;j<rowNum;j++){
18                 boolean valid = false;
19                 for (int k=0;k<colNum;k++)
20                     if (board[j][k]==word.charAt(0)){
21                         boolean[][] visited = new boolean[rowNum][colNum];
22                         for (int p = 0;p<rowNum;p++)
23                             Arrays.fill(visited[p],false);
24                         valid = isValidWord(board,visited,word,0,j,k);
25                         if (valid){
26                             res.add(word);
27                             break;
28                         }
29                     }
30                 if (valid) break;
31             }
32         }
33 
34         return res;
35     }
36     
37     public boolean isValidWord(char[][] board, boolean[][] visited, String word, int pos, int x, int y){
38         if (x<0 || x>=board.length || y<0 || y>=board[0].length) return false;        
39         if (word.charAt(pos)!=board[x][y] || visited[x][y]) return false;
40 
41         if (pos==word.length()-1)
42                 return true;
43         
44 
45         visited[x][y] = true;
46         if (isValidWord(board,visited,word,pos+1,x+1,y) || isValidWord(board,visited,word,pos+1,x-1,y) || isValidWord(board,visited,word,pos+1,x,y+1) || isValidWord(board,visited,word,pos+1,x,y-1))
47             return true;
48         else {
49             visited[x][y]=false;
50             return false;
51         }
52     }
53 }

 

LintCode-Word Search II

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原文地址:http://www.cnblogs.com/lishiblog/p/4183802.html

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