79. Word Search

https://leetcode.com/problems/word-search/description/
在一个二维字符矩阵中搜索是否存在给定字符串，要求只能水平或垂直搜索
思路：回溯。对每个位置遍历一遍，分别对（i，j）的上下左右位置递归。

第一种写法，对一般的测试用例能通过，但对于特别大的矩阵会报超时。

class Solution:
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        self.length = 0  #记录匹配的串长
        #遍历整个矩阵
        for i in range(len(board)):
            for j in range(len(board[0])):
                if word[0] == board[i][j]:  #找到切入点
                    board[i][j] = '.'  #标记为已访问
                    self.backtrack(board, word, 1, i, j)  #，从word[1]开始
                    board[i][j] = word[0]  #还原现场
                if self.length > 0:
                    return True
        return False

    #回溯函数，count为word中已找到的字符数，row、col为矩阵坐标
    def backtrack(self, board, word, count, row, col):
        if count == len(word):  #整个串都找到
            self.length = count
        else:
            #上
            if row - 1 >= 0 and board[row - 1][col] == word[count]:
                board[row - 1][col] = '.'
                self.backtrack(board, word, count + 1, row - 1, col)
                board[row - 1][col] = word[count]
            #下
            if row + 1 < len(board) and board[row + 1][col] == word[count]:
                board[row + 1][col] = '.'
                self.backtrack(board, word, count + 1, row + 1, col)
                board[row + 1][col] = word[count]
            #左
            if col - 1 >= 0 and board[row][col - 1] == word[count]:
                board[row][col - 1] = '.'
                self.backtrack(board, word, count + 1, row, col - 1)
                board[row][col - 1] = word[count]
            #右
            if col + 1 < len(board[0]) and board[row][col + 1] == word[count]:
                board[row][col + 1] = '.'
                self.backtrack(board, word, count + 1, row, col + 1)
                board[row][col + 1] = word[count]

改进写法：主要改进递归函数，很多操作，例如将某个位置置‘.’和恢复现场，是可以归纳到一起的。此外也不用额外寻找‘切入点’，归纳到一起即可。

class Solution:
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        self.length = 0
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.backtrack(board, word, 0, i, j):  #从word[0]开始
                    return True
        return False

    def backtrack(self, board, word, count, row, col):
        if count == len(word):
            return True
        #下标越界，或者当前位置字符不匹配
        if row < 0 or row >= len(board) or col < 0 or col >= len(board[0]) or board[row][col] != word[count]:
            return False
        #防止重复访问
        board[row][col] = '.'
        #对四个方向递归搜索，注意是or
        res = self.backtrack(board, word, count+1, row-1, col) or self.backtrack(board, word, count+1, row+1, col) or self.backtrack(board, word, count+1, row, col-1) or self.backtrack(board, word, count+1, row, col+1)
        #还原现场
        board[row][col] = word[count]
        return res