[Leetcode 1014] Best Sightseeing Pair

原题说明

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

解题思路

遍历数组A，我们用max_score来记录当前最大的分数，用pre_max来记录当前元素之前最高的分数
每遍历一个元素a，我们首先更新max_score: max_score = max(max_score, a + pre_max)
然后更新pre_max: pre_max = max(pre_max, a) - 1
这里-1是因为每往后一个元素，距离会增加1

示例代码 (cpp)

class Solution {
public:
    int maxScoreSightseeingPair(vector<int>& A) {
        int max_score = 0, pre_max = 0;
        for (auto a : A) {
            max_score = max(max_score, a + pre_max);
            pre_max = max(pre_max, a) - 1;
        }
        return max_score;
    }
};

示例代码（java）

class Solution {
    public int maxScoreSightseeingPair(int[] A) {
        int res = 0, cur = 0;
        for (int a: A) {
            res = Math.max(res, cur + a);
            cur = Math.max(cur, a) - 1;
        }
        return res;
    }
};

示例代码（python）

class Solution(object):
    def maxScoreSightseeingPair(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        cur = res = 0
        for a in A:
            res = max(res, cur + a)
            cur = max(cur, a) - 1
        return res

复杂度分析

时间复杂度: O(n)
空间复杂度: O(1)

归纳总结

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