leetcode1306

 1 class Solution:
 2     def bfs(self,arr,n,start,visited,path):
 3         while len(path) > 0:
 4             temp = []
 5             while len(path) > 0:
 6                 idx = path.pop()
 7                 for i in range(n):
 8                     if visited[i] == 1 or i == idx:
 9                         continue
10                     else:
11                         if i < idx and arr[i] == idx - i:
12                             if i == start:
13                                 return True
14                             visited[i] = 1
15                             temp.append(i)
16                         elif i > idx and arr[i] == i - idx:
17                             if i == start:
18                                 return True
19                             visited[i] = 1
20                             temp.append(i)
21             path = temp[:]
22         return False
23 
24 
25     def canReach(self, arr: 'List[int]', start: int) -> bool:
26         if arr[start] == 0:
27             return True
28         path = []#记录0的下标
29         n = len(arr)
30         visited = [0] * n
31         for i in range(n):
32             if arr[i] == 0:
33                 path.append(i)
34         return self.bfs(arr,n,start,visited,path)

Algorithm thinking: BFS.

First record index 0 all the values ​​of the elements, as the initial set. Use two to traverse, to find out the set up of each and every point into the next layer.

Use visited cache, filtered repeated point visited, preventing ring die loops.