Description

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.

分析

题目的意思是：给你一个数组，找出最大的sightseeing pair对，即求下面的最大值：

A[i] + A[j] + i - j

这道题我暴力破解了一下，果然超时了，后面参考了一下别人的实现，global_max存放的就是遍历过程中的全局最大值，也是最后的结果，max_val存放的是第一个值，ma_inx就是对应的索引了。这里如果要更新max_val的话，要满足当前的值A[i]>=max_val+max_inx-i就行了，如果没有想到这个点就做不出来了。

代码

class Solution:
    def maxScoreSightseeingPair(self, A: List[int]) -> int:
        n=len(A)
        max_val=A[0]
        max_inx=0
        global_max=-1
        for i in range(1,n):
            if(A[i]+max_val+max_inx-i>global_max):
                global_max=A[i]+max_val+max_inx-i
            if(max_val+max_inx-i<=A[i]):
                max_val=A[i]
                max_inx=i
        return global_max

参考文献

​​[LeetCode] Easy Python solution beats 98%​​