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 3Example 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 3Example 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 * 10^40 <= arr[i] < arr.length0 <= start < arr.length
跳跃游戏III。题意是给你一个数组，和一个起点start，是数组中的一个下标，请你返回是否可以经过数次跳跃，到达指定下标的位置。可以往左或者往右跳跃，但是不允许越界。
思路是BFS。这道题跟前两个版本几乎没有任何联系。当你在某一个下标start的时候，因为可以往左也可以往右跳跃，所以需要判断跳跃之后的index是否越界，如果不越界，则放入queue进行下一轮循环。从queue中弹出元素的时候，记得需要判断这个点是否遍历过，如果没有被遍历过，最后也需要加入hashset记录一下。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 public boolean canReach(int[] arr, int start) { 3 int len = arr.length; 4 HashSet<Integer> visited = new HashSet<>(); 5 Queue<Integer> queue = new LinkedList<>(); 6 queue.offer(start); 7 while (!queue.isEmpty()) { 8 int i = queue.poll(); 9 if (arr[i] == 0) { 10 return true; 11 } 12 if (visited.contains(i)) { 13 continue; 14 } 15 visited.add(i); 16 if (i + arr[i] < len) { 17 queue.offer(i + arr[i]); 18 } 19 if (i - arr[i] >= 0) { 20 queue.offer(i - arr[i]); 21 } 22 } 23 return false; 24 } 25 }