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English Version

题目描述

给你一个整数数组 nums 和一个整数 k ,请你返回 非空 子序列元素和的最大值,子序列需要满足:子序列中每两个 相邻 的整数 nums[i] 和 nums[j] ,它们在原数组中的下标 i 和 j 满足 i < j 且 j - i <= k

数组的子序列定义为:将数组中的若干个数字删除(可以删除 0 个数字),剩下的数字按照原本的顺序排布。

 

示例 1:

输入:nums = [10,2,-10,5,20], k = 2
输出:37
解释:子序列为 [10, 2, 5, 20] 。

示例 2:

输入:nums = [-1,-2,-3], k = 1
输出:-1
解释:子序列必须是非空的,所以我们选择最大的数字。

示例 3:

输入:nums = [10,-2,-10,-5,20], k = 2
输出:23
解释:子序列为 [10, -2, -5, 20] 。

 

提示:

  • 1 <= k <= nums.length <= 10^5
  • -10^4 <= nums[i] <= 10^4

解法

方法一:动态规划 + 单调队列

class Solution:
    def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
        n = len(nums)
        dp = [0] * n
        ans = -inf
        q = deque()
        for i, v in enumerate(nums):
            if q and i - q[0] > k:
                q.popleft()
            dp[i] = max(0, 0 if not q else dp[q[0]]) + v
            while q and dp[q[-1]] <= dp[i]:
                q.pop()
            q.append(i)
            ans = max(ans, dp[i])
        return ans
class Solution {
    public int constrainedSubsetSum(int[] nums, int k) {
        int n = nums.length;
        int[] dp = new int[n];
        int ans = Integer.MIN_VALUE;
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            if (!q.isEmpty() && i - q.peek() > k) {
                q.poll();
            }
            dp[i] = Math.max(0, q.isEmpty() ? 0 : dp[q.peek()]) + nums[i];
            while (!q.isEmpty() && dp[q.peekLast()] <= dp[i]) {
                q.pollLast();
            }
            q.offer(i);
            ans = Math.max(ans, dp[i]);
        }
        return ans;
    }
}
class Solution {
public:
    int constrainedSubsetSum(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> dp(n);
        int ans = INT_MIN;
        deque<int> q;
        for (int i = 0; i < n; ++i) {
            if (!q.empty() && i - q.front() > k) q.pop_front();
            dp[i] = max(0, q.empty() ? 0 : dp[q.front()]) + nums[i];
            ans = max(ans, dp[i]);
            while (!q.empty() && dp[q.back()] <= dp[i]) q.pop_back();
            q.push_back(i);
        }
        return ans;
    }
};
func constrainedSubsetSum(nums []int, k int) int {
	n := len(nums)
	dp := make([]int, n)
	ans := math.MinInt32
	q := []int{}
	for i, v := range nums {
		if len(q) > 0 && i-q[0] > k {
			q = q[1:]
		}
		dp[i] = v
		if len(q) > 0 && dp[q[0]] > 0 {
			dp[i] += dp[q[0]]
		}
		for len(q) > 0 && dp[q[len(q)-1]] < dp[i] {
			q = q[:len(q)-1]
		}
		q = append(q, i)
		ans = max(ans, dp[i])
	}
	return ans
}