题目如下：

Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [4,2,3,0,3,1,2], start = 5
Output: true
Explanation: 
All possible ways to reach at index 3 with value 0 are: 
index 5 -> index 4 -> index 1 -> index 3 
index 5 -> index 6 -> index 4 -> index 1 -> index 3 

Example 2:

Input: arr = [4,2,3,0,3,1,2], start = 0
Output: true 
Explanation: 
One possible way to reach at index 3 with value 0 is: 
index 0 -> index 4 -> index 1 -> index 3

Example 3:

Input: arr = [3,0,2,1,2], start = 2
Output: false
Explanation: There is no way to reach at index 1 with value 0.

Constraints:

• 1 <= arr.length <= 5 * 10^4
• 0 <= arr[i] < arr.length
• 0 <= start < arr.length

解题思路：DFS。从起点开始，每次把可以到达的位置标记成可达到即可。

代码如下：

class Solution(object):
    def canReach(self, arr, start):
        """
        :type arr: List[int]
        :type start: int
        :rtype: bool
        """
        reach = [0] * len(arr)
        queue = [start]
        reach[start] = 1
        while len(queue) > 0:
            inx = queue.pop(0)
            if arr[inx] == 0 and reach[inx] == 1:
                return True
            next_inx = inx + arr[inx]
            if next_inx >= 0 and next_inx < len(arr) and reach[next_inx] == 0:
                queue.append(next_inx)
                reach[next_inx] = 1
            next_inx = inx - arr[inx]
            if next_inx >= 0 and next_inx < len(arr) and reach[next_inx] == 0:
                queue.append(next_inx)
                reach[next_inx] = 1
        return False