题目：原题链接（中等）
标签：回溯算法、深度优先搜索
| 解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
|---|---|---|---|
| Ans 1 (Python) | O ( ( M N ) 2 ) O((MN)^2) O((MN)2) | O ( M × N ) O(M×N) O(M×N) | 1964ms (13.10%) |
| Ans 2 (Python) | |||
| Ans 3 (Python) |
解法一：
class Solution:
def getMaximumGold(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
def _is_valid(x, y):
return 0 <= x < m and 0 <= y < n
def _get_neighbors(x1, y1):
return [(x2, y2) for (x2, y2) in [(x1 - 1, y1), (x1 + 1, y1), (x1, y1 - 1), (x1, y1 + 1)]
if _is_valid(x2, y2)]
waiting = set()
for i in range(m):
for j in range(n):
if grid[i][j] > 0:
waiting.add((i, j))
ans = 0
for i1, j1 in waiting:
def dfs(ii1, jj1, vv1):
res = vv1
for ii2, jj2 in _get_neighbors(ii1, jj1):
vv2 = vv1 + grid[ii2][jj2]
if grid[ii2][jj2] > 0 and (ii2, jj2) not in visited:
visited.add((ii2, jj2))
res = max(res, dfs(ii2, jj2, vv2))
visited.remove((ii2, jj2))
return res
visited = {(i1, j1)}
ans = max(ans, dfs(i1, j1, grid[i1][j1]))
return ans
