LeetCode(79):Word Search

Word Search: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 =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

题意：在一个给定的矩阵中查找给定的单词是否出现，可以向上、下、左、右进行查找。不能在一个字母上重复查找。

思路：采用深度有限搜索和递归的方式进行解决。

代码：

public class Solution {
    public boolean exist(char[][] board, String word) {
        if(word.length()==0){return false;}
        boolean[][] visited = new boolean[board.length][board[0].length];
        for(int i=0;i<board.length;i++){
            for(int j=0;j<board[i].length;j++){
                visited[i][j] = false;
            }
        }
        
        for(int i=0;i<board.length;i++){
            
            for(int j=0;j<board[i].length;j++){
                if(board[i][j]==word.charAt(0)){
                    visited[i][j] = true;
                    if(word.length()==1||search(board,i,j,word.substring(1),visited)){
                        return true;
                    }
                    visited[i][j] = false;
                }
            }
        }
        return false;
    }
    boolean search(char[][] board,int i,int j,String word,boolean[][] visited){
        if(word.length()==0){return true;}
        int[][] direction={{-1,0},{1,0},{0,-1},{0,1}};
        for(int k=0;k<direction.length;k++){
            int ii = i + direction[k][0];
            int jj = j + direction[k][1];
            if(ii>=0&&ii<board.length&&jj>=0&&jj<board[i].length&&board[ii][jj]==word.charAt(0)&&visited[ii][jj]==false){
                visited[ii][jj]=true;
                if(word.length()==1||search(board,ii,jj,word.substring(1),visited)){
                    return true;
                }
                visited[ii][jj]=false;
            }
        }
            return false;
        }
}