leetcode 1219. Path with Maximum Gold 解法 python
一.问题描述
In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.
Return the maximum amount of gold you can collect under the conditions:
- Every time you are located in a cell you will collect all the gold in that cell.
- From your position you can walk one step to the left, right, up or down.
- You can't visit the same cell more than once.
- Never visit a cell with
0gold. - You can start and stop collecting gold from any position in the grid that has some gold.
Example 1:
Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
Output: 24
Explanation:
[[0,6,0],
[5,8,7],
[0,9,0]]
Path to get the maximum gold, 9 -> 8 -> 7.
Example 2:
Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
Output: 28
Explanation:
[[1,0,7],
[2,0,6],
[3,4,5],
[0,3,0],
[9,0,20]]
Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.
Constraints:
1 <= grid.length, grid[i].length <= 150 <= grid[i][j] <= 100- There are at most 25 cells containing gold.
二.解决思路
这种题目比较明显要用深度优先搜索来解决，尝试当前路径后记录当前路径下的值，然后回溯到上一层
具体来说：遍历grid，对每一个cell值不是0的进行dfs，记录值
具体的dfs的话请自行百度
我比较喜欢用递归解决所有东西，所以我实现的dfs不是那么标准
当然最好还是用标准的dfs，写算法题和真正应用不一样，标准dfs也可以帮我们理清思路
主要要注意的地方有两点：
第一：注意python的深拷贝和浅拷贝
第二：在某条路径下递归完后记得把isvisited给重新设置成false，不然会影响其他方向上的迭代
更新：自己还是写了个稍微标准点的，不知为啥最后TLE？
最后看了一下可能问题出在第二个版本我用了集合来追踪cell的状态，但是集合操作如add和remove很花时间？
还有其他想法的欢迎评论区交流或者私信！
建议直接看第二个版本的，第二个版本逻辑清楚一些
更多leetcode算法题解法请关注我的专栏leetcode算法从零到结束或关注我
欢迎大家一起套路一起刷题一起ac
三.源码
import copy
class Solution:
def getMaximumGold(self, grid: List[List[int]]) -> int:
maxnum=-1
is_visited=[[False for j in range(len(grid[0]))] for i in range(len(grid))]
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j]!=0:
temp=copy.deepcopy(is_visited)
temp[i][j]=True
count=self.searchPath(i,j,grid,temp,grid[i][j])
maxnum=count if count>maxnum else maxnum
return maxnum
def searchPath(self, i,j,grid,is_visited,count):
num1,num2,num3,num4=count,count,count,count
if i-1>=0 and grid[i-1][j] and not is_visited[i-1][j]:
is_visited[i-1][j]=True
num1=self.searchPath(i-1,j,grid,is_visited,count+grid[i-1][j])
is_visited[i-1][j]=False
if i+1=0 and grid[i][j-1] and not is_visited[i][j-1]:
is_visited[i][j-1]=True
num3=self.searchPath(i,j-1,grid,is_visited,count+grid[i][j-1])
is_visited[i][j-1]=False
if j+1 class Solution:
def getMaximumGold(self, grid: List[List[int]]) -> int:
is_visited=set()
return max(self.DFS(i,j,grid,is_visited) for j in range(len(grid[0])) \
for i in range(len(grid)) if grid[i][j]!=0)
def DFS(self, i,j,grid,is_visited):
if i<0 or i>=len(grid) or j<0 or j>=len(grid[0]) or (i,j) in is_visited \
or grid[i][j]==0:
return -1
is_visited.add((i,j))
rst=grid[i][j]+max(0,max(self.DFS(i+x,j+y,grid,is_visited) \
for x,y in [[-1,0],[1,0],[0,-1],[0,1]]))
is_visited.remove((i,j))
return rst