[Swift]LeetCode1014. 最佳观光组合 | Best Sightseeing Pair

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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:

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

---

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。

示例：

输入：[8,1,5,2,6]
输出：11
解释：i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11

提示：

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

---

Runtime: 404 ms

Memory Usage: 19.7 MB

class Solution {
    func maxScoreSightseeingPair(_ A: [Int]) -> Int {
        var n:Int = A.count
        var best:Int = -Int.max
        var most:Int = -Int.max
        for i in 0..<n
        {
            best = max(best, A[i] - i + most)
            most = max(most, A[i] + i)
        }
        return best       
    }
}

---

404ms

class Solution {
    func maxScoreSightseeingPair(_ A: [Int]) -> Int {
        var maxScore = Int.min
        var leftHighScoreIndex = 0
        var leftHighScoreValue = A[leftHighScoreIndex]
        for i in 1..<A.count {
            maxScore = max(leftHighScoreValue + A[i] + leftHighScoreIndex - i, maxScore)
            if (A[i] - leftHighScoreValue) + (i - leftHighScoreIndex) >= 0 {
                leftHighScoreIndex = i 
                leftHighScoreValue = A[i]
            }
        }
        return maxScore
    }
}

---

440ms

class Solution {
    func maxScoreSightseeingPair(_ A: [Int]) -> Int {
        // B: store max of A[0]+0 ~ A[i]+i
        var B = A, currMax = B[0] + 0
        for i in 0..<B.count {
            currMax = max(B[i] + i, currMax)
            B[i] = currMax
        }
        
        var dp = A.map { _ in 0 }
        for i in 1..<dp.count {
            dp[i] = max(dp[i-1], A[i] - i + B[i-1])
        }
        
        return dp.last!
    }
}

---

460ms

class Solution {
    func maxScoreSightseeingPair(_ A: [Int]) -> Int {
        var score = 0
        
        var s1 = Array<Int>()
        s1.reserveCapacity(A.count)
      
        for i in 0..<A.count {
            s1.append(A[i] + i)
        }
        
        var m = Array<Int>()
        m.append(A.last! - (A.count - 1))
        
        for i in (1..<A.count) {
           m.append(max(A[A.count - i - 1] - (A.count - i - 1), m[i - 1]))
        }
        m.reverse()
      
        for i in 0..<(A.count-1) {
            score = max(score, s1[i] + m[i + 1])
        }
        
        
        return score
    }
}

---

464ms

class Solution {
    func maxScoreSightseeingPair(_ A: [Int]) -> Int {
        var tmp = Int.min, result = Int.min
        for (offset, element) in A.enumerated() {
            result = max(result, element - offset + tmp)
            tmp = max(element + offset, tmp)
        }
        
        return result
    }
}

---

524ms

class Solution {
    func maxScoreSightseeingPair(_ A: [Int]) -> Int {
        if A.count < 2 { return -1 }        
        var cache = [Int]()        
        cache.append(0)        
        var res = Int.min        
        for idx in 1 ..< A.count {            
            res = max(res, A[cache.last!] + A[idx] + cache.last! - idx)
            if A[idx] - cache.last! + idx > A[cache.last!] {
                cache.append(idx)
            }
        }        
        return res        
    }
}