LeetCode79 单词搜索

题目

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

 提示： 
 m == board.length 
 n = board[i].length 
 1 <= m, n <= 6 
 1 <= word.length <= 15 
 board 和 word 仅由大小写英文字母组成 

方法

剪枝法

• 时间复杂度：一个非常宽松的上界为 O(MN⋅3^L),其中 M, N 为网格的长度与宽度，LL 为字符串word 的长度。每个方向只有一次，只能走3次
• 空间复杂度：O(MN)。我们额外开辟了 O(MN) 的 visited 数组，同时栈的深度最大为O(min(L,MN))。

class Solution {
    public boolean exist(char[][] board, String word) {
        if(board==null||board.length<1) return false;
        int row = board.length,col = board[0].length;
        boolean[][] visited = new boolean[row][col];
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                boolean res = dfs(board,i,j,word,0,visited);
                if(res) return true;
            }
        }
        return false;
    }
    private boolean dfs(char[][] board,int i,int j,String word,int index,boolean[][] visited){
        if(board[i][j]!=word.charAt(index)) {
            return false;
        }
        if(index==word.length()-1){
            return true;
        }
        visited[i][j] = true;
        int[][] dict = {{1,0},{-1,0},{0,1},{0,-1}};
        boolean result = false;
        for(int[] d:dict){
            int new_i = i+d[0],new_j = j+d[1];
            if((new_i>=0&&new_i<board.length&&new_j>=0&&new_j<board[0].length)&&!visited[new_i][new_j]){
                boolean res = dfs(board,new_i,new_j,word,index+1,visited);
                if(res){
                    result = true;
                    break;
                }
            }
        }
        visited[i][j] = false;
        return result;
    }
}