【c++刷题Day2】专题3栈与队列&单调栈与单调队列

这是C++刷题的Day2

🕋题目描述
🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀
An array of size n ≤ 106 is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.
Window position Minimum value Maximum value
[1 3 -1] -3 5 3 6 7 -1 3
1 [3 -1 -3] 5 3 6 7 -3 3
1 3 [-1 -3 5] 3 6 7 -3 5
1 3 -1 [-3 5 3] 6 7 -3 5
1 3 -1 -3 [5 3 6] 7 3 6
1 3 -1 -3 5 [3 6 7] 3 7

Your task is to determine the maximum and minimum values in the sliding window at each position.
1

Input
The input consists of two lines. The first line contains two integers n and k which are the lengths of the array and the sliding window. There are n integers in the second line.
Output
There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.
Sample
Inputcopy Outputcopy

8 3
1 3 -1 -3 5 3 6 7

	

-1 -3 -3 -3 3 3
3 3 5 5 6 7
1
2
3
4
5
6
7

🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀🚀

🔑思路

使用单调队列，维护最大值最小值：
实时的维护一个单调的内部序列，最后队首就是当前的最小值&最大值

💯CODE代码：

#include 
#pragma GCC optimize(2)
#define min3 (x , y , z) min (x , min (y , z))
#define max3 (x , y , z) max (x , max (y , z))

using namespace std ;

typedef long long LL ;
typedef double LF ;
typedef pair < int , int > PII ;
typedef pair < string , int > PSI ;
const int maxn = 1e6 + 10 ;
int q[maxn] , a[maxn] , n , k ;

int main () {


    scanf ("%d%d" , & n , & k) ;
    for (int i = 0; i < n; i++)
        scanf ("%d" , & a[i]) ;
    int hh = 0 , tt = -1 ;
    for (int i = 0; i < n; i++) {
        if (hh <= tt && i - k + 1 > q[hh])
            hh ++ ;
        while (hh <= tt && a[q[tt]] >= a[i])
            tt -- ;
        q[++ tt] = i ;
        if (i >= k - 1)
            printf ("%d " , a[q[hh]]) ;
    }
    puts ("") ;
    hh = 0 , tt = -1 ;
    for (int i = 0; i < n; i++) {
        if (hh <= tt && i - k + 1 > q[hh])
            hh ++ ;
        while (hh <= tt && a[q[tt]] <= a[i])
            tt -- ;
        q[++ tt] = i ;
        if (i >= k - 1)
            printf ("%d " , a[q[hh]]) ;
    }


    return 0 ;
}
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