zoukankan      html  css  js  c++  java
  • 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:

      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)

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    hiho #1502:最大子矩阵(元素和不超过k)
    IPC 进程间通信方式——消息队列
    IPC 进程间通信方式——共享内存
    IPC 进程间通信方式——管道
    hiho #1032: 最长回文子串
    TCP超时与重传机制与拥塞避免
    C++关于构造函数 和 析构函数 能否抛出异常的讨论
    基于TCP的客户端、服务器端socket编程
    hiho #1043 : 完全背包
    hiho #1485 : hiho字符串(滑动窗口)
  • 原文地址:https://www.cnblogs.com/h-hkai/p/12822819.html
Copyright © 2011-2022 走看看