1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit

Given an array of integers nums and an integer limit, return the size of the longest continuous subarray such that the absolute difference between any two elements is less than or equal to limit.

In case there is no subarray satisfying the given condition return 0.

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 <= 10^5
• 1 <= nums[i] <= 10^9
• 0 <= limit <= 10^9

题意：

　　给出一个数组，求连续子串的最大长度，所求的子串要求满足 |max_element - min_element| < limit

思路：

　　用两个双向队列来模拟，max_deque用来存储当前所在位置之前元素的最大值，min_deque用来存储当前所在位置之前所有元素的最小值。用两个指针leftPointer和rightPointer来寻找最长的subArray，通过对rightPointer向后迭代，更新max_deque 和 min_deque，同时通过shrink leftPointer来使subArray满足题目要求。

Code：

class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        deque<int> max_deque, min_deque;
        int left = 0, ans = 0;
        for (int right = 0; right < nums.size(); ++right) {
            while (!max_deque.empty() && max_deque.back() < nums[right])
                max_deque.pop_back();
            while (!min_deque.empty() && min_deque.back() > nums[right])
                min_deque.pop_back();
            max_deque.push_back(nums[right]);
            min_deque.push_back(nums[right]);
            while (max_deque.front() - min_deque.front() > limit) {
                if (max_deque.front() == nums[left]) max_deque.pop_front();
                if (min_deque.front() == nums[left]) min_deque.pop_front();
                left++;
            }
            ans = max(ans, right - left + 1);
        }
        return ans;
    }
};

参考：

　　https://leetcode.com/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/discuss/609743/Java-Detailed-Explanation-Sliding-Window-Deque-O(N)