[LeetCode] Best Sightseeing Pair

Given an array A of positive integers, A[i] represents the value of the i-th sightseeing spot, and two sightseeing spots i and j have distance j - i between them.

The score of a pair (i < j) of sightseeing spots is (A[i] + A[j] + i - j) : the sum of the values of the sightseeing spots, minus the distance between them.

Return the maximum score of a pair of sightseeing spots.

Example 1:

Input: [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11

Note:

2 <= A.length <= 50000
1 <= A[i] <= 1000

解题思路

维护两个状态转移方程：

• res = max(res, cur+a)
• cur = max(cur, a) - 1

cur表示曾经遇到过的最大得分，但是它每走一步都会衰减1，res当前到当前的最大得分。

实现代码

// Runtime: 4 ms, faster than 67.70% of Java online submissions for Best Sightseeing Pair.
// Memory Usage: 50.3 MB, less than 100.00% of Java online submissions for Best Sightseeing Pair.
class Solution {
    public int maxScoreSightseeingPair(int[] A) {
        int res = 0, cur = 0;
        for (int a : A) {
            res = Math.max(res, cur + a);
            cur = Math.max(cur, a) - 1;
        }
        return res;
    }
}