[一起来刷leetcode吧][32]--No.79 Integer to Roman-优快云博客

本文介绍LeetCode第79题的解决方案，该题要求在一个二维矩阵中查找给定单词是否存在。通过递归搜索的方法实现，避免重复使用同一字母。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

这篇文章是程序自动发表的,详情可以见这里
href="http://ounix1xcw.bkt.clouddn.com/github.markdown.css" rel="stylesheet">

这是leetcode的第79题--Integer to Roman

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.

思路这道题可以用递归函数来解决，注意到，eg从“a” 中判断是否有“aa” ，如果直接递归是返回true的，这不对，要就是说在一次判断中，已判断过的字母不能再判断，所以参数应该还有一个board，记录当前的字母状态，考虑到python是用的引用，传递参数时没有复制，我又不想额外再去复制board，就在函数中记下当前位置的字母，在函数调用结束再改回来的。哈哈！

show me the code

class Solution(object):
    def exist(self, board, word):
        """
        :type board: List[List[str]]
        :type word: str
        :rtype: bool
        """
        row = len(board)
        col = len(board[0])
        num = len(word)
        def find(r,c,n):
            if n == num-1 and board[r][c] == word[n]:
                return True
            if board[r][c] != word[n] or n == num-1:
                return  False
            tmp = board[r][c]    #save  the  val
            board[r][c] = 0
            if r < row-1 and find(r 1,c,n 1):
                return True
            if r > 0 and find(r-1,c,n 1):
                return True
            if c <col-1 and find(r,1 c,n 1):
                return True
            if c > 0 and find(r,c-1,n 1):
                return True
            board[r][c] = tmp
            return False
        for i in range(row):
            for j in range(col):
                if find(i,j,0):
                    return True
        return False