LeetCode 솔루션 분류
[11/24] 79. Word Search
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Medium
12096487Add to ListShareGiven an m x n grid of characters board and a string word, return true if word exists in the grid.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" Output: true
Example 2:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" Output: true
Example 3:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" Output: false
Constraints:
m == board.lengthn = board[i].length1 <= m, n <= 61 <= word.length <= 15boardandwordconsists of only lowercase and uppercase English letters.
Follow up: Could you use search pruning to make your solution faster with a larger board?
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class Solution:
def exist(self, board: List[List[str]], word: str) -> bool:
ROW, COL = len(board), len(board[0])
visited = set()
def dfs(row, col, i):
if (row >= ROW or col >= COL or row < 0 or col < 0 or (row,col) in visited or word[i] != board[row][col]):
return False
visited.add((row,col))
if i == len(word)-1:
return True
ret_val = dfs(row+1, col, i+1) or dfs(row, col+1, i+1) or dfs(row-1,col,i+1) or dfs(row, col-1, i+1)
visited.remove((row,col))
return ret_val
for row in range(ROW):
for col in range(COL):
if dfs(row, col, 0): return True
return False
