LeetCode1499. Max Value of Equation——单调队列

一、题目

You are 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.

Return 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.

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 <= 105
points[i].length == 2
-108 <= xi, yi <= 108
0 <= k <= 2 * 108
xi < xj for all 1 <= i < j <= points.length
xi form a strictly increasing sequence.

二、题解

class Solution {
public:
    int findMaxValueOfEquation(vector<vector<int>>& points, int k) {
        int n = points.size(),res = INT_MIN;
        deque<int> q;
        for(int i = 0;i < n;i++){
            int x = points[i][0];
            int y = points[i][1];
            while(!q.empty() && points[q.front()][0] + k < x) q.pop_front();
            if(!q.empty()){
                res = max(res,x + y + points[q.front()][1] - points[q.front()][0]);
            }
            while(!q.empty() && y - x >= points[q.back()][1] - points[q.back()][0]){
                q.pop_back();
            }
            q.push_back(i);
        }
        return res;
    }
};