1423.Maximum Points You Can Obtain from Cards

Tags: Medium Sliding Windows Dynamic Programming

Links: https://leetcode.com/problems/maximum-points-you-can-obtain-from-cards/

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There are several cards arranged in a row, and each card has an associated number of points The points are given in the integer array cardPoints.

In one step, you can take one card from the beginning or from the end of the row. You have to take exactly k cards.

Your score is the sum of the points of the cards you have taken.

Given the integer array cardPoints and the integer k, return the maximum score you can obtain.

Example 1:

Input: cardPoints = [1,2,3,4,5,6,1], k = 3
Output: 12
Explanation: After the first step, your score will always be 1. However, choosing the rightmost card first will maximize your total score. The optimal strategy is to take the three cards on the right, giving a final score of 1 + 6 + 5 = 12.

Example 2:

Input: cardPoints = [2,2,2], k = 2
Output: 4
Explanation: Regardless of which two cards you take, your score will always be 4.

Example 3:

Input: cardPoints = [9,7,7,9,7,7,9], k = 7
Output: 55
Explanation: You have to take all the cards. Your score is the sum of points of all cards.

Example 4:

Input: cardPoints = [1,1000,1], k = 1
Output: 1
Explanation: You cannot take the card in the middle. Your best score is 1. 

Example 5:

Input: cardPoints = [1,79,80,1,1,1,200,1], k = 3
Output: 202

Constraints:

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

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第一眼看到这个题，联想到了《挑战程序设计竞赛》2.2.3字典序最小问题的例题，POJ 3617 Best Cow Line，贪心的方法，但是很容易举出测试用例来否定贪心的方法：

3 2 1 100 1
k = 2

那么换一种思路，究竟在在i次选择（$1 \leq i \leq k$）从需要首部取还是尾部取得顺序，对于最终结果无影响，但是必须满足假如从首部取走n张卡片，那么这n张必须是连续的。连续的子数组，就可以联想到双指针的解法。

第一次遍历，从头开始计算前k个数的和，然后用一个指针end指向序列尾端的后一个位置，代表最开始全部从首部选择，然后end逐渐向首部移动，每次移动一个单位，当end移动k个单位的时候，代表此时全部从尾部取。

那么其实只需要用一个变量sum来维护从首部取得元素得总和，每次减去一个元素，再用tail维护尾部的元素总和，每次增加一个元素。

时间复杂度$O(n)$，空间复杂度$O(n)$.

class Solution {
public:
    int maxScore(vector<int>& cardPoints, int k) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = cardPoints.size();
        int sum = 0;
        for (int i = 0; i < k; ++i) sum += cardPoints[i];
        int maxVal = sum;

        int end = n;
        int tail = 0;
        for (int i = k - 1; i >= 0; --i) {
            sum -= cardPoints[i];
            tail += cardPoints[--end];

            maxVal = max(maxVal, sum + tail);
        }

        return maxVal;
    }
};