Leetcode 1425: DP + 单调队列

纸上得来终觉浅绝知此事要躬行2023-09-23 13:57

Constrained Subsequence Sum
Hard

Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums $i$ and nums $j$ , where i < j, the condition j - i <= k is satisfied.

A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order.

Example 1:

Input: nums = $10,2,-10,5,20$ , k = 2

Output: 37

Explanation: The subsequence is $10, 2, 5, 20$ .

Example 2:

Input: nums = $-1,-2,-3$ , k = 1

Output: -1

Explanation: The subsequence must be non-empty, so we choose the largest number.

Example 3:

Input: nums = $10,-2,-10,-5,20$ , k = 2

Output: 23

Explanation: The subsequence is $10, -2, -5, 20$ .

Constraints:

1 <= k <= nums.length <= 105

-104 <= nums $i$ <= 104

解法1：DP。会超时。

cpp 复制代码

class Solution {
public:
    int constrainedSubsetSum(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> dp(n, INT_MIN / 3);
        dp[0] = nums[0];
        int res = dp[0];
        for (int i = 1; i < n; i++) {
            dp[i] = nums[i];
            for (int j = 1; j <= k; j++) {
                
                if (i >= j) {
                    dp[i] = max(dp[i], dp[i - j] + nums[i]);
                }       
            }
            res = max(res, dp[i]);
        }
       
        return res;
    }
};

解法2：DP+单调队列。注意单调队列本身就是一个滑动窗口。

cpp 复制代码

class Solution {
public:
    int constrainedSubsetSum(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> dp(n, INT_MIN / 3);
        dp[0] = nums[0];
        int res = dp[0];
        deque<int> dq;
        dq.push_back(0);
        for (int i = 1; i < n; i++) {
            while (!dq.empty() && i - dq.front() > k) dq.pop_front(); 
            dp[i] = max(nums[i], dp[dq.front()] + nums[i]);
            while (!dq.empty() && dp[dq.back()] < dp[i]) dq.pop_back();
            dq.push_back(i);
            res = max(res, dp[i]);
        }
        return res;
    }
};