Leetcode79 Word Search

题目描述

board =[  ['A','B','C','E'],  
          ['S','F','C','S'],  
          ['A','D','E','E']]​
Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

思路

根据word的第一个字符(d=0)在二维char数组中匹配，找到起始位置。找到起始位置后，在该位置的左右上下search, 直到找到与word.charAt(d+1)相同的元素。当到达边界以后（即到达最后一行或最后一列）时返回false，当没有找到word.charAt(d+1)匹配的board元素时，返回false. 当d到达word的最后一个元素后，返回true. 设定boolean型变量flag, flag的值为上下左右search或的结果。
keep in mind：search过的board元素应当暂时改为0.

代码

class Solution {
    public boolean exist(char[][] board, String word) {
        if(board.length==0) return false;
        for(int x=0;x<board.length;x++){//row
            for(int y=0;y<board[0].length;y++){//columns
                if(search(board,word,x,y,0))
                    return true;
            }
        }
        return false;
    }
    
    private boolean search(char[][] board, String word, int x, int y, int d){
        if(x<0||x>=board.length||y<0||y>=board[0].length)
            return false;
        if(word.charAt(d)!=board[x][y])
            return false;
        if(d==word.length()-1)
            return true;
        char pre = board[x][y];
        board[x][y]=0;
        boolean flag = search(board,word,x+1,y,d+1)||search(board,word,x-1,y,d+1)
            ||search(board,word,x,y+1,d+1)||search(board,word,x,y-1,d+1);
        board[x][y]=pre;
        return flag;
    }
}