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.
Constraints:
1 <= cardPoints.length <= 10^51 <= cardPoints[i] <= 10^41 <= k <= cardPoints.length
题目链接：https://leetcode.com/problems/maximum-points-you-can-obtain-from-cards/
题目大意：有n个数字，每次只能从两边取，一共可以取k个，求能取出的数字和的最大值
题目分析：求前缀和，然后枚举从左边取i个(0 <= i <= k)时总共能取到的最大值
3ms，时间击败53.43%
class Solution {
public int maxScore(int[] cardPoints, int k) {
int n = cardPoints.length;
int[] sum = new int[n + 1];
for (int i = 1; i <= n; i++) {
sum[i] = sum[i - 1] + cardPoints[i - 1];
}
int ans = 0;
for (int i = 0; i <= k; i++) {
ans = Math.max(ans, sum[i] + sum[n] - sum[n - k + i]);
}
return ans;
}
}
