Leetcode 79. Word Search

https://leetcode.com/problems/word-search/

Medium

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.

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.

---

• 经典回溯。注意Python列表初始化，复习DFS模版。

 

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        if not word:
            return False
        
        visited = [ [ False for j in range( len( board[ 0 ] ) ) ] for i in range( len( board ) ) ] # remember the way
        
        def helper(x, y, current):
            if current == len( word ):
                return True
            
            if x < 0 or x >= len( board ) or y < 0 or y >= len( board[ 0 ] ) or visited[x][y] or board[x][y] != word[current]:
                return False
            
            visited[x][y] = True
            
            result = (helper(x - 1, y, current + 1) 
                    or helper(x + 1, y, current + 1) 
                    or helper(x, y - 1, current + 1) 
                    or helper(x, y + 1, current + 1) )
            
            visited[x][y] = False
            
            return result
        
        for i in range( len( board ) ):
            for j in range( len ( board[ 0 ] ) ):
                if helper(i, j, 0):
                    return True
        
        return False