LeetCode 79. Word Search

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