题目描述

Given an array points containing the coordinates of points on a 2D plane, sorted by the x-values, where points[i] = [xi, yi] such that xi < xj for all 1 <= i < j <= points.length. You are also given an integer k.

Find the maximum value of the equation yi + yj + |xi - xj| where |xi - xj| <= k and 1 <= i < j <= points.length. It is guaranteed that there exists at least one pair of points that satisfy the constraint |xi - xj| <= k.

样例

Example 1:

Input: points = [[1,3],[2,0],[5,10],[6,-10]], k = 1
Output: 4
Explanation: The first two points satisfy the condition |xi - xj| <= 1 and if we calculate the equation we get 3 + 0 + |1 - 2| = 4. Third and fourth points also satisfy the condition and give a value of 10 + -10 + |5 - 6| = 1.
No other pairs satisfy the condition, so we return the max of 4 and 1.

Example 2:

Input: points = [[0,0],[3,0],[9,2]], k = 3
Output: 3
Explanation: Only the first two points have an absolute difference of 3 or less in the x-values, and give the value of 0 + 0 + |0 - 3| = 3.

Constraints:

• 2 <= points.length <= 10^5
• points[i].length == 2
• -10^8 <= points[i][0], points[i][1] <= 10^8
• 0 <= k <= 2 * 10^8
• points[i][0] < points[j][0] for all 1 <= i < j <= points.length
• xi form a strictly increasing sequence.

算法

(单调队列) $O(n)$

因为题目保证数据xi < xj，所以不等式变形为(yi - xi) + xj + yj，于是就变成了找当前点前面横坐标差值不超过k的最大的yi - xi，很明显的单调队列的模型。

单调队列里每个元素存储两个数据，第一个是横坐标xi，第二个是yi - xi，队列里维护区间里yi - xi的最大值。

时间复杂度$O(n)$。

C++ 代码

class Solution {
public:
    int findMaxValueOfEquation(vector<vector<int>>& points, int k) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        deque<vector<int>> dq;
        int res = INT_MIN;
        for (const auto & e : points) {
            while (!dq.empty() && e[0] - dq.front()[0] > k) dq.pop_front();
            if (!dq.empty()) res = max(res, e[0] + e[1] + dq.front()[1]);
            while (!dq.empty() && dq.back()[1] < e[1] - e[0]) dq.pop_back();
            dq.push_back(vector<int>{e[0], e[1] - e[0]});
        }

        return res;
    }
};