給定一個?m x n 二維字符網格?board 和一個字符串單詞?word 。如果?word 存在于網格中，返回 true ；否則，返回 false 。單詞必須按照字母順序，通過相鄰的單元格內的字母構成，其中“相鄰”單元格是那些水平相鄰或垂直相鄰的單元格。同一個單元格內的字母不允許被重復使用。示例 1：輸入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" 輸出：true 示例 2：輸入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" 輸出：true 示例 3：輸入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" 輸出：false提示：m == board.length n = board[i].length 1 <= m, n <= 6 1 <= word.length <= 15 board 和 word 僅由大小寫英文字母組成鏈接：https://leetcode-cn.com/problems/word-searchclass Solution:def exist(self, board: List[List[str]], word: str) -> bool:row, col = len(board), len(board[0])def dfs(x, y, index):if board[x][y] != word[index]:return Falseif index == len(word) - 1:return Trueboard[x][y] = '#'for c in [[0,1],[0,-1],[-1,0],[1,0]]:nx, ny = x + c[0], y + c[1]if 0 <= nx < row and 0 <= ny < col and dfs(nx, ny, index+1):return Trueboard[x][y] = word[index]for i in range(row):for j in range(col):if dfs(i,j,0):return Truereturn False

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