1014. Best Sightseeing Pair

[Tcdian]

1014. Best Sightseeing Pair

给定正整数数组 A，A[i] 表示第 i 个观光景点的评分，并且两个景点 i 和 j 之间的距离为 j - i。

一对景点（i < j）组成的观光组合的得分为（A[i] + A[j] + i - j）：景点的评分之和减去它们两者之间的距离。

返回一对观光景点能取得的最高分。

Example

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

[Tcdian]

Solution

• JavaScript Solution

/**
 * @param {number[]} A
 * @return {number}
 */
var maxScoreSightseeingPair = function(A) {
    let result = -Infinity;
    let maxScoreSightseeingPoint = A[0] + 0;
    for (let i = 1; i < A.length; i++) {
        result = Math.max(maxScoreSightseeingPoint + A[i] - i, result);
        maxScoreSightseeingPoint = Math.max(A[i] + i, maxScoreSightseeingPoint);
    }
    return result;
};

• TypeScript Solution

function maxScoreSightseeingPair(A: number[]): number {
    let result = -Infinity;
    let maxScoreSightseeingPoint = A[0] + 0;
    for (let i = 1; i < A.length; i++) {
        result = Math.max(maxScoreSightseeingPoint + A[i] - i, result);
        maxScoreSightseeingPoint = Math.max(A[i] + i, maxScoreSightseeingPoint);
    }
    return result;
};