leetcode-记忆化深搜/动态规划v2

leetcode-记忆化深搜/动态规划v1_林冲风雪山神庙的博客-CSDN博客

741. 摘樱桃

我的错误解法

class Solution(object):
    def cherryPickup(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        #两个人同时从起点摘樱桃
        def dfs(x1,y1,x2,y2):
            if x1 >= len(grid) or y1 >= len(grid) or x2 >= len(grid) or y2 >= len(grid):
                return float("-inf")
            if grid[x1][y1] == -1 or grid[x2][y2] == -1:
                return float("-inf")
            if (x1,y1) == (x2,y2):
                res = grid[x1][y1]
            else:
                res = grid[x1][y1] + grid[x2][y2]
            res += max(dfs(x1+1,y1,x2,y2),dfs(x1,y1+1,x2,y2))
            res += max(dfs(x1,y1,x2+1,y2),dfs(x1,y1,x2,y2+1))
            return res
        return dfs(0,0,0,0)

class Solution:
    def cherryPickup(self, grid: List[List[int]]) -> int:
        @lru_cache(None)
        def dfs(x1,y1,x2,y2):
            if x1 > len(grid)-1 or x2 > len(grid)-1 or y1 > len(grid[0])-1 or y2 > len(grid[0])-1:
                return float('-inf')
            if grid[x1][y1] == -1 or grid[x2][y2] == -1:
                return float('-inf')
            # # 到终点
            if x1 == y1 == x2 == y2 == len(grid)-1:
                return grid[x1][y1]
            res = 0
            if (x1,y1) == (x2,y2):
                res += grid[x1][y1]
            else:
                res += grid[x1][y1] + grid[x2][y2]
            #从两个点的移动方向中找最大的，两个人同时移动
            res += max(dfs(x1+1,y1,x2+1,y2),dfs(x1,y1+1,x2,y2+1),dfs(x1+1,y1,x2,y2+1),dfs(x1,y1+1,x2+1,y2))
            return res 
        res = dfs(0,0,0,0)
        if res == float("-inf"):
            return 0
        return res

174. 地下城游戏