[leetcode]79. Word Search找单词

佚名 6年前 (2019-05-16) 随笔 1714人围观抢沙发百度已收录

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.

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Example:

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.

题意：

给定一个字母矩阵，允许从任一点出发四处走。看轨迹能否构成一个单词。

 1 class Solution {
 2     public boolean exist(char[][] board, String word) {
 3         for(int i = 0; i< board.length; i++){
 4             for(int j = 0; j < board[0].length; j++){
 5                 if(helper(word, 0, board, i, j)){
 6                     return true;
 7                 }
 8             }
 9         }
10         return false;
11     }
12     
13     private boolean helper(String word, int index, char[][]board, int i, int j){
14         // base case
15         if(index == word.length()){
16             return true;
17         }
18         // 1. out of bound  2. char doesn't match
19         if( i < 0 || i >= board.length || j < 0 || j >= board[0].length || word.charAt(index) != board[i][j]){
20             return false;
21         }
22         
23         // save current character data and make a flag here 
24         char temp = board[i][j];
25         board[i][j] = '0';
26         boolean found = helper (word, index+1, board, i+1, j) 
27                         ||helper (word, index+1, board, i-1, j) 
28                         ||helper (word, index+1, board, i, j-1) 
29                         ||helper (word, index+1, board, i, j+1); 
30        // dismiss flag
31        board[i][j] = temp;
32        return found;         
33     }  
34 }