【leetcode】1696. Jump Game VI

题目如下：

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

解题思路：动态规划，要求跳到第i个位置的最大值，只需要找出i-1~i-k区间的最大值即可。

代码如下：

class Solution(object):
    def maxResult(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        import bisect
        dp = [0] * len(nums)
        dp[0] = nums[0]

        val = [dp[0]]

        for i in range(1,len(nums)):
            dp[i] = val[-1] + nums[i]
            bisect.insort_left(val,dp[i])
            if len(val) > k:
                inx = bisect.bisect_left(val,dp[i-k])
                del val[inx]
        #print dp
        return dp[-1]