[LeetCode] 1306. Jump Game III

轉自LeetCode

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 * 104
• 0 <= arr[i] < arr.length
• 0 <= start < arr.length

Solution

這題可以用 BFS 的想法思考

滿像就是在迷宮裡面，能不能從入口走到出口，只要找到一個答案就結束

所以用個 queue  儲存合法可以走的 index，同時也用個 set 來記錄走過的路

如果都沒找到，就回傳 false 即可

kotlin

	class Solution {
	fun canReach(arr: IntArray, start: Int): Boolean {
	val queue = mutableListOf<Int>()
	queue.add(start)
	var nextIndex = 0
	val set = mutableSetOf<Int>()
	while(queue.isNotEmpty()) {
	val index = queue.first()
	queue.removeAt(0)
	if(arr[index] == 0) {
	return true
	}
	nextIndex = index + arr[index]
	if(nextIndex < arr.size && !set.contains(nextIndex)) {
	queue.add(nextIndex)
	set.add(nextIndex)
	}
	nextIndex = index - arr[index]
	if(nextIndex > -1 && !set.contains(nextIndex)) {
	queue.add(nextIndex)
	set.add(nextIndex)
	}
	}
	return false
	}
	}

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