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:
1 class Solution {
2 public:
3 int longestSubarray(vector<int>& nums, int limit) {
4 deque<int> max_deque, min_deque;
5 int left = 0, ans = 0;
6 for (int right = 0; right < nums.size(); ++right) {
7 while (!max_deque.empty() && max_deque.back() < nums[right])
8 max_deque.pop_back();
9 while (!min_deque.empty() && min_deque.back() > nums[right])
10 min_deque.pop_back();
11 max_deque.push_back(nums[right]);
12 min_deque.push_back(nums[right]);
13 while (max_deque.front() - min_deque.front() > limit) {
14 if (max_deque.front() == nums[left]) max_deque.pop_front();
15 if (min_deque.front() == nums[left]) min_deque.pop_front();
16 left++;
17 }
18 ans = max(ans, right - left + 1);
19 }
20 return ans;
21 }
22 };
参考:
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