※ 引述《DJYOSHITAKA (franchouchouISBEST)》之銘言： : 原本想要寫第二題 : 被你版系列文搞到:( : py好難 : 我一定有一堆地方寫的有問題 : 懶得檢查 對不起 : def dfs(self, i, j, cur, cur_max, grid: List[List[int]]) -> int: : if self.visit[i][j] == 1 or grid[i][j] == 0: : return cur : m = len(grid) : n = len(grid[0]) : self.visit[i][j] = 1 : cur += grid[i][j] : if i-1 >= 0: : cur_max = max(cur_max, self.dfs(i-1, j, cur, cur_max, grid)) : if i+1 < m: : cur_max = max(cur_max, self.dfs(i+1, j, cur, cur_max, grid)) : if j-1 >= 0: : cur_max = max(cur_max, self.dfs(i, j-1, cur, cur_max, grid)) : if j+1 < n: : cur_max = max(cur_max, self.dfs(i, j+1, cur, cur_max, grid)) : self.visit[i][j] = 0 : return cur_max : def getMaximumGold(self, grid: List[List[int]]) -> int: : ans = 0 : for i in range(len(grid)): : for j in range(len(grid[0])): : self.visit = [[0]*len(grid[0]) for i in range(len(grid))] : ans = max(ans, self.dfs(i, j, 0, 0, grid)) : return ans 思路: 經典DFS Python Code: class Solution: def getMaximumGold(self, grid: List[List[int]]) -> int: n = len(grid) m = len(grid[0]) max_res = 0 def dfs(grid,visited,x,y): visited.add((x,y)) if grid[y][x] == 0: return 0 else: res = grid[y][x] dx = [1,-1,0,0] dy = [0,0,1,-1] for i in range(4): new_x = x+dx[i] new_y = y+dy[i] if new_x>=0 and new_x<m and new_y>=0 and new_y<n: if grid[new_y][new_x] !=0 and (new_x,new_y) not in visited: res = max(dfs(grid,visited,new_x,new_y)+grid[y][x],res) visited.remove((x,y)) return res for i in range(n): for j in range(m): max_res = max(dfs(grid,set(),j,i),max_res) return max_res 晚安捏 -- ※ 發信站: 批踢踢實業坊(ptt-website.tw), 來自: 114.43.138.16 (臺灣) ※ 文章網址: https://ptt-website.tw/Marginalman/M.1715710635.A.423

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