LeetCode1423. 可获得的最大点数 [Medium]

1423. Maximum Points You Can Obtain from Cards

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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题目：从数组的两端（每次操作可从任意一端）取k个数，使得这k个数的和最大。

思路：滑窗，参考link。要找到满足和最大的数组两端的k个数，则意味着剩下的n-k个数的和最小，且该n-k个数必然在数组中连续，因此以n-k的大小进行滑窗判断。

工程代码下载

class Solution {
public:
    int maxScore(vector<int>& cardPoints, int k) {
        int n = cardPoints.size();
        int total = 0, min_score = INT_MAX;
        for(int i = 0, j = 0, cur = 0; j < n; ++j){
            cur += cardPoints[j];
            total += cardPoints[j];
            if(j - i + 1 > n - k)
                cur -= cardPoints[i++];
            // 注意两个if并列，不用else if，针对 n - k == 0 的情况
            if(j - i + 1 == n - k)
                min_score = min(min_score, cur);
        }
        return total - min_score;
    }
};