【leetcode】1021. Best Sightseeing Pair

题目如下：

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 <= 50000

1 <= A[i] <= 1000

解题思路：本题和【leetcode】123. Best Time to Buy and Sell Stock III 有点类似。题目要求求出A[i] + A[j] + i - j最大值，这个表达式可以等价变为 (A[i] + i ) + (A[j] - j），对于A中每一个元素来说，都对应着两个值，分别是A[i]+i和A[i]-i。本题的解法可以抽象成把A分成两部分，第一部分找出使得A[i]+i能得到最大值的i，第二部分找出A[j]-j能得到最大值的j，两者之和就是最终的结果。

代码如下：

class Solution(object):
    def maxScoreSightseeingPair(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        l_1 = [-float('inf')] * len(A)
        l_2 = [-float('inf')] * len(A)
        max_1 = -float('inf')
        max_2 = -float('inf')
        res = -float('inf')
        for i in range(len(A)-1):
            max_1 = max(i + A[i],max_1)
            l_1[i] = max_1

            max_2 = max(A[len(A) - 1-i] - (len(A) - 1-i),max_2)
            l_2[len(A) - 1-i] = max_2

            if i >= len(A) / 2 -1:
                res = max(res,l_1[i] + l_2[i+1])
                res = max(res,l_1[len(A) - 1-i-1] + l_2[len(A) - 1-i])
        #print l_1
        #print l_2
        return res