Leetcode1438 Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit (滑动窗口+双单调队列)

1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit
      Medium
      Given an array of integers nums and an integer limit, return the size of the longest non-empty subarray such that the absolute difference between any two elements of this subarray is less than or equal to limit.

Example 1:

Input: nums = [8,2,4,7], limit = 4
Output: 2
Explanation: All subarrays are:
[8] with maximum absolute diff |8-8| = 0 <= 4.
[8,2] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4] with maximum absolute diff |8-2| = 6 > 4.
[8,2,4,7] with maximum absolute diff |8-2| = 6 > 4.
[2] with maximum absolute diff |2-2| = 0 <= 4.
[2,4] with maximum absolute diff |2-4| = 2 <= 4.
[2,4,7] with maximum absolute diff |2-7| = 5 > 4.
[4] with maximum absolute diff |4-4| = 0 <= 4.
[4,7] with maximum absolute diff |4-7| = 3 <= 4.
[7] with maximum absolute diff |7-7| = 0 <= 4.
Therefore, the size of the longest subarray is 2.
Example 2:

Input: nums = [10,1,2,4,7,2], limit = 5
Output: 4
Explanation: The subarray [2,4,7,2] is the longest since the maximum absolute diff is |2-7| = 5 <= 5.
Example 3:

Input: nums = [4,2,2,2,4,4,2,2], limit = 0
Output: 3

Constraints:

1 <= nums.length <= 105
1 <= nums[i] <= 109
0 <= limit <= 109

解法1：这题我觉得不简单。
要维护一个滑动窗口和两个单调队列(一个维护窗口最大值，一个维护窗口最小值)。
注意:

1. 这两个单调队列(maxQ和minQ)里面的值不一样，但窗口的边界(left, right)是一致的。也就是说，滑动窗口就一个，里面的最小值就是minQ的最小值(front)，里面的最大值就是maxQ的最大值(front)。
2. maxQ和minQ的back就是滑动窗口的右边界(right)，注意因为right每次加1，所以maxQ和minQ的back可能是right-1。
3. maxQ和minQ的front可能会比滑动窗口的左边界(left)大一些，当然也可能等于left，但不会小于left。也就是说，滑动窗口的left, right边界会完全cover住maxQ和minQ。
4. minQ是单调递增队列(从front到back单调递增)，maxQ是单调递减队列(从front到back单调递减)。minQ的front就是滑动窗口的最小值，maxQ的front就是滑动窗口的最大值。
5. 每次进nums[right]，maxQ和minQ根据各自的back来决定要不要pop back。
6. 当maxQ的front - minQ的front 对应的值的difference > limit的时候，我们就要决定移动left。直到这两者的difference < limit。注意：maxQ的front和minQ的front有可能>left, 这时候我们不能pop_front，因为它们必须保留在滑动窗口内。只有front<=left的时候，才能安全pop_front()。

class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        int n = nums.size();
        deque<int> minQ, maxQ;
        int left = 0, right = 0, res = 0;
        while (right < n) {
            while(minQ.size() > 0 && nums[minQ.back()] > nums[right]) {
                minQ.pop_back();
            }
            minQ.push_back(right);
            
            while(maxQ.size() > 0 && nums[maxQ.back()] < nums[right]) {
                maxQ.pop_back();
            }
            maxQ.push_back(right);
            right++;
           
            while (nums[maxQ.front()] - nums[minQ.front()] > limit) {
                if (maxQ.front() <= left) maxQ.pop_front();
                if (minQ.front() <= left) minQ.pop_front();
                left++;
            }
            res = max(res, right - left);
        }
        return res;
    }
};