1423. 可获得的最大点数

• 从数组两边取，则中间剩余部分连续，则考虑使用滑动窗口
• 窗口中值最小，对应的是头尾取值的最大值
• 剩余元素个数 n- k，即滑动窗口大小

class Solution {
public:
    int maxScore(vector<int>& cardPoints, int k) {
        int n = cardPoints.size();
        int total = accumulate(cardPoints.begin(), cardPoints.end(), 0);
        int win = n - k;
        int winsum = accumulate(cardPoints.begin(), cardPoints.begin() + win, 0);
        int minsum = winsum;
        for(int i = win; i < n; i++)
        {
            winsum += cardPoints[i] - cardPoints[i - win];
            minsum = min(winsum, minsum);
        }
        return total - minsum;
    }
};