标签：pair not values maximum distance ret out exit imu
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 <= 500001 <= A[i] <= 1000Idea 1. max(A[i] + A[j] + i -j) (i < j) = max(max(A[i] + i) + A[j] -j), scan from left to right
Time complexity: O(n)
Space complexity: O(1)
1 class Solution { 2 public int maxScoreSightseeingPair(int[] A) { 3 int maxScore = Integer.MIN_VALUE; 4 int sumVal = Integer.MIN_VALUE; 5 for(int i = 1; i < A.length; ++i) { 6 sumVal = Math.max(sumVal, A[i-1] + (i-1)); 7 maxScore = Math.max(maxScore, A[i] - i + sumVal); 8 } 9 10 return maxScore; 11 } 12 }
Idea 1.b scan from right to left, max(A[i] + A[j] + i -j) (i < j) = max(max(A[j] -j) + A[i] + i)
1 class Solution { 2 public int maxScoreSightseeingPair(int[] A) { 3 int maxScore = Integer.MIN_VALUE; 4 int sumVal = Integer.MIN_VALUE; 5 for(int i = A.length-2; i >= 0; --i) { 6 sumVal = Math.max(sumVal, A[i+1] - (i+1)); 7 maxScore = Math.max(maxScore, A[i] + i + sumVal); 8 } 9 10 return maxScore; 11 } 12 }
标签：pair not values maximum distance ret out exit imu
原文地址：https://www.cnblogs.com/taste-it-own-it-love-it/p/10652767.html
