Sunday, October 4, 2020

1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit ---M !!!!~~~

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


A:

这个题目错了在2个地方。

第一首先是没有看清题意。  忽略了any .

************method 1 ， use order sorted data structure ********

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        int curMax = 0;
        // use multiset to save the sorted values
        multiset<int> MS; // default is decending order
        int start = 0;
        for(int i =0;i<nums.size();i++){
            MS.insert(nums[i]);
            while(*MS.rbegin() - *MS.begin() > limit){
                MS.erase(MS.find(nums[start++]));
            }
            curMax = max(curMax, int(MS.size()) );
        }
        return curMax;
    }
};

Mistakes:

multiset 不能直接 *Mset.end() 因为其值未定义。  所以要用Mset.rbegin()


*****************method 2 , using Deque to save the MAX and MIN so far ***************

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public:
    int longestSubarray(vector<int>& nums, int limit) {
        int n = nums.size();
        int start = 0;
        int res = 0;
        deque<int> increaseD, decreaseD;
        for(int i =0;i<n; i++){
            int val = nums[i];
            while(not increaseD.empty() && increaseD.back() > val){
                increaseD.pop_back();
            }
            increaseD.push_back(val);
            while((not decreaseD.empty() && decreaseD.back() < val )){
                decreaseD.pop_back();
            }
            decreaseD.push_back(val);
            while(decreaseD.front() - increaseD.front() > limit){
                int delVal = nums[start++];
                if(increaseD.front() == delVal ){
                    increaseD.pop_front();
                }
                if(decreaseD.front() == delVal){
                    decreaseD.pop_front();
                }
            }
            res = max(res, i - start+1);
        }
        return res;
    }
};


Mistakes:

line 18,  decreaseD.front() - increaseD.front()  写反了。 哎，SB了 


Learned:

1:  using multiset . default multiset<int>  is descending order






at

No comments:

Post a Comment