1696. Jump Game VI - kumaeki/LeetCode GitHub Wiki
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 <= 10^5
- -10^4 <= nums[i] <= 10^4
Dynamic Programming. 但是需要做一些优化, 不然时间复杂度是N(nk).
class Solution {
public int maxResult(int[] nums, int k) {
int n = nums.length;
int[] memo = new int[n];
memo[0] = nums[0];
ArrayDeque<Integer> que = new ArrayDeque<Integer>(k);
que.offer(0);
for (int i = 1; i < n; i++){
if(que.getFirst() < i - k)
que.removeFirst();
memo[i] = nums[i] + memo[que.getFirst()];
// 保证que中不存在比当前memo[i]小的值 的下标
// 并且从左到右总是递减的顺序排列
while(!que.isEmpty() && memo[que.getLast()] <= memo[i])
que.removeLast();
que.addLast(i);
}
return memo[n - 1];
}
}