Word Search

package cn.edu.xidian.sselab.array;
/**
 * title: Word Search
 * content:
 *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.
 */
public class WordSearch {

    //这个题就像是玩贪吃蛇的游戏，抓住重点，关键点就是如何构造递归函数，以及什么时候返回值
    //这个题目还有一个难点是，每个点不能重复使用，也就是说每个判断过正确的点，判断下一个节点的时候，这个值不能用，可以随便赋个值
    //只要保证他的判断返回是false就可以
    public boolean exist(char[][] board, String word) {
        int totalLen = board.length;
        int sigleLen = board[0].length;
        for(int i=0;i<totalLen;i++){
            for(int j=0;j<sigleLen;j++){
                if(exists(board,i,j,word,0)){
                    return true;
                }
            }
        }
        return false;
   }
    
    private boolean exists(char[][]board,int i,int j,String word,int temp){
        if(temp == word.length())
            return true;
        if(i<0 || j<0 || i == board.length|| j == board[0].length || board[i][j] != word.charAt(temp))
            return false;
        board[i][j] = '*';
        boolean result = exists(board,i-1,j,word,temp+1)||
                 exists(board,i+1,j,word,temp+1)||
                 exists(board,i,j-1,word,temp+1)||
                 exists(board,i,j+1,word,temp+1);
        board[i][j] = word.charAt(temp);
        return result;
    }

}