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  • 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
        }
    }
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  • 原文地址:https://www.cnblogs.com/johnnyzhao/p/14175456.html
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