LeetCode 79. Word Search

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本文介绍了一种使用回溯法在二维字符网格中查找指定单词的方法。通过递归地检查每个可能路径，确保同一字符不被重复使用。文章提供了一个Python实现示例。

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题目：

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.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,

word = "ABCB", -> returns false.

思路：利用一个flag数组标记已走过的点,遇到已走过的点就返回试下一条路，利用回溯法递归可求解如下。

class Solution(object):
def exist(self,board, word):
"""
:type board: List[List[str]]
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
:type word: str
:rtype: bool
"""
if not word or len(word)==0:return True
rows = len(board);columns = len(board[0])
print board
visited = [[False for j in range(columns)] for i in range(rows)]
for i in range(rows):
for j in range(columns):
if board[i][j]==word[0]:
visited[i][j]=True
if self.dfs(board,visited,word[1:],i,j):
return True
visited[i][j]=False
return False

def dfs(self,board,visited,word,x_new,y_new):
if word=='':return True
ctrl = [(0,-1),(0,1),(1,0),(-1,0)]
for i,j in ctrl:
x = i+x_new;y=j+y_new
if (0<=x<len(board)) and (0<=y<len(board[0])) and (word[0]==board[x][y]) and (visited[x][y]==False):
visited[x][y]=True
if self.dfs(board,visited,word[1:],x,y):
return True
visited[x][y]=False
return False