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