1423. 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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解法1

分别计算出顺序和倒序依次的累积sum(running sum), 然后遍历所有可能性得到最大值(sliding window)

class Solution {
    public int maxScore(int[] cardPoints, int k) {
        int len = cardPoints.length;
        // 定义两个数组, 一个顺序保存sum, 一个倒序保存sum
        // 注意数组长度是len + 1, 顺序的第一个和倒序的最后一个表示空
        int[] arrAsc = new int[len + 1], arrDesc = new int[len + 1];
        for(int i = 1; i <= len; i++){
            // 顺序
            arrAsc[i] = arrAsc[i - 1] + cardPoints[i - 1];
            // 倒序
            arrDesc[len - i] = arrDesc[len - i + 1] + cardPoints[len - i];
        }
        
        int res = 0;
        // 计算每种可能的组合,得到最大值
        for(int i = 0 ; i <= k ; i++)
            res = Math.max(res, arrAsc[k - i] + arrDesc[len - i]);

        return res;
    }
}