1014. Best Sightseeing Pair

package LeetCode_1014

/**
 * 1014. Best Sightseeing Pair
 * https://leetcode.com/problems/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:
1. 2 <= A.length <= 50000
2. 1 <= A[i] <= 1000
 * */
class Solution {
    /*
    * solution: by rule: A[i] + A[j] + i - j, change into: (A[i]+i)+(A[j]-j),
    * so we need to get: max(A[i]+i) and max(A[j]-j),
    * Time:O(n), Space:(1)
    * */
    fun maxScoreSightseeingPair(A: IntArray): Int {
        var max = Int.MIN_VALUE
        //max A[i]+i
        var maxI = Int.MIN_VALUE
        //max A[j]-j
        var maxJ = Int.MIN_VALUE
        val n = A.size
        for (i in 0 until n - 1) {
            if (A[i] + i > maxI) {
                maxI = A[i] + i
                //if maxI got the new value, because j depend on i, need reset maxJ
                maxJ = Int.MIN_VALUE
            }
            //because i < j, set j=i+1
            val j = i + 1
            maxJ = Math.max(maxJ, A[j] - j)
            max = Math.max(max, maxI + maxJ)
        }
        return max
    }
}
原文地址:https://www.cnblogs.com/johnnyzhao/p/14175456.html