[LeetCode] 1696. Jump Game VI

轉自LeetCode

You are given a 0-indexed integer array nums and an integer k.

You are initially standing at index 0. In one move, you can jump at most k steps forward without going outside the boundaries of the array. That is, you can jump from index i to any index in the range [i + 1, min(n - 1, i + k)] inclusive.

You want to reach the last index of the array (index n - 1). Your score is the sum of all nums[j] for each index j you visited in the array.

Return the maximum score you can get.

 

Example 1:

Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7
Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.

Example 2:

Input: nums = [10,-5,-2,4,0,3], k = 3
Output: 17
Explanation: You can choose your jumps forming the subsequence [10,4,3] (underlined above). The sum is 17.

Example 3:

Input: nums = [1,-5,-20,4,-1,3,-6,-3], k = 2
Output: 0

 

Constraints:

• 1 <= nums.length, k <= 105
• -104 <= nums[i] <= 104

Solution

這題和239. Sliding Window Maximum 可以說同個思路

這題還需要將 slicing window 和 DP 兩個概念結合在一起

一樣用個 stack 來記錄目前的最大值的 index

而一個關鍵點在，把原本的 nums 拿來當 DP 用，每次得到的當下最大值，都更新到 nums

kotlin

	class Solution {
	fun maxResult(nums: IntArray, k: Int): Int {
	val stack = mutableListOf<Int>()
	var currValue = 0
	for(i in nums.indices) {
	if(stack.isNotEmpty() && stack.first() == i - k - 1) {
	stack.removeAt(0)
	}
	currValue = if(stack.isEmpty()) nums[i] else nums[i] + nums[stack.first()]
	while(stack.isNotEmpty() && currValue > nums[stack.last()]) {
	stack.removeAt(stack.lastIndex)
	}
	nums[i] = currValue
	stack.add(i)
	}
	return currValue
	}
	}

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