leetcode刷题日记: 79. 单词搜索（java）

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true
示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true
示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/word-search
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
    //一共四个方向
    int [] dx = new int[]{0,0,1,-1};
    int [] dy = new int[]{1,-1,0,0};
    public boolean exist(char[][] board, String word) {
        for(int i =0;i<board.length;i++){
            for(int j=0;j<board[i].length;j++){
                if (dfs(board,word,0,i,j)){
                    return true;
                }
            }
        }
        return false;
    }
    public boolean dfs(char[][] board,String word,int index,int x ,int y){
        if(board[x][y]!=word.charAt(index)){
            return false;
        }
        if(index == word.length()-1){
            return true;
        }
        char temp = board[x][y];
        //设置为访问状态
        board[x][y] = '.';
        for(int i =0;i<4;i++){
            int new_x = x+dx[i];
            int new_y = y+dy[i];
            //处理边界条件
            if(new_x<0 ||new_x> board.length-1 ||new_y<0||new_y>board[0].length-1||board[new_x][new_y] == '.'){
                continue;
            }
            if(dfs(board,word,index+1,new_x,new_y)) return true;
        }
        //访问完成之后进行还原
        board[x][y] = temp;
        return false;

    }
}

标签：word,java,int,单元格,board,new,true,leetcode,79
来源： https://blog.csdn.net/qq_42038061/article/details/120743912