79. 单词搜索
给定一个二维网格和一个单词,找出该单词是否存在于网格中。
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:
board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
给定 word = "ABCCED", 返回 true.
给定 word = "SEE", 返回 true.
给定 word = "ABCB", 返回 false.
解题思路:以board数组的任意元素为起始点,进行深度优先搜索(回溯算法),若搜索出单词,则返回True;若以任意元素为起始点均不能搜索出单词,返回False。
Python3代码如下:
class Solution(object):
def exist(self, board, word):
"""
:type board: List[List[str]]
:type word: str
:rtype: bool
"""
m,n = len(board),len(board[0])
visited = [[0] * n for i in range(m)]
for i in range(m):
for j in range(n):
if self.dfs(0,i,j,visited,board,word):
return True
return False
def dfs(self,curr,i,j,visited,board,word):
if curr == len(word):
return True
if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or board[i][j] != word[curr] or visited[i][j]:
return False
visited[i][j] = 1
result = self.dfs(curr+1,i+1,j,visited,board,word) or self.dfs(curr+1,i-1,j,visited,board,word) or self.dfs(curr+1,i,j+1,visited,board,word) or self.dfs(curr+1,i,j-1,visited,board,word)
visited[i][j] = 0
return result