leetcood学习笔记-79-单词搜索
题目描述:
方法一;回溯
class Solution: def exist(self, board: List[List[str]], word: str) -> bool: max_x,max_y,max_step = len(board)-1,len(board[0])-1,len(word)-1 def maze(x, y,step,visited): if visited[x][y]==1: return False if board[x][y] != word[step]: return False if step==max_step: return True visited[x][y]=1 if x < max_x and maze(x+1,y,step+1,visited): return True if x>0 and maze(x-1,y,step+1,visited): return True if y<max_y and maze(x,y+1,step+1,visited): return True if y>0 and maze(x,y-1,step+1,visited): return True # 记得失败后要置零 visited[x][y]=0 return False visited = [[0]*(max_y+1) for i in range(max_x+1)] for x in range(max_x+1): for y in range(max_y+1): if board[x][y] != word[0]: continue if maze(x,y,0,visited): return True return False
优化:
class Solution: def exist(self, board: List[List[str]], word: str) -> bool: if not board: return False row = len(board) col = len(board[0]) def dfs(word, i, j, visited): if not word: return True for li, lj in [[0,1],[1,0],[-1,0],[0,-1]]: tmp_i = i + li tmp_j = j + lj if 0 <= tmp_i < row and 0 <= tmp_j < col and word[0] == board[tmp_i][tmp_j] and (tmp_i,tmp_j) not in visited: visited.add((tmp_i,tmp_j)) if dfs(word[1:], tmp_i,tmp_j, visited): return True visited.remove((tmp_i,tmp_j)) return False for i in range(row): for j in range(col): if board[i][j] == word[0] and dfs(word[1:], i, j,{(i,j)} ): return True return False