单调栈和单调队列

输入样例：

5
3 4 2 7 5

输出样例：

-1 3 -1 2 2

普通做法,时间复杂度O(n^2),可能会TLE

#include 
using namespace std;

const int N = 1e5+10;
int st[N];

int main()
{
    int n,x;
    int res;
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &st[i]);
        int ret = 0;
        for(int j = i-1; j >= 0; j --){
            if(st[j] < st[i]){
                ret = 1;
                res = st[j];
                break;
            }
        }
        if(ret) printf("%d ",res);
        else printf("-1 ");
        
    }
    return 0;
}

单调栈,时间复杂度O(n)

保证每次输出时,栈顶均为该位置前的最小数

(与上一种做法相比,去除了大量重复且无意义的步骤,每个数只会进出栈一次,而上一种做法双重循环会使得很多数被反复遍历) 

#include 
using namespace std;

const int N = 1e5+10;
int st[N],top;//栈

int main()
{
    int n,x;
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &x);
        
        while(top && st[top]>=x) top--;
        if(!top) printf("-1 ");
        else printf("%d ",st[top]);
        
        st[++top] = x;
    }
    return 0;
}

154. 滑动窗口 - AcWing题库高质量的算法题库https://www.acwing.com/problem/content/description/156/

不断进出队列, 保证输出的时候队列的队头存着当前窗口最小/最大值 在数组a中的下标

#include 
using namespace std;

const int N = 1e6+10;
int a[N],q[N];

int main()
{
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
    
    int h = 0, t = -1;//h-队头下标, t-队尾下标
    for (int i = 0; i < n; i ++ )
    {
        if(h <= t && i+1-k > q[h]) h++;
        while(h <= t && a[i] <= a[q[t]]) t--;
        
        q[++t] = i;
        if(i >= k-1) printf("%d ",a[q[h]]);
    }
    cout << endl;
    
    h = 0, t = -1;
    for (int i = 0; i < n; i ++ )
    {
        if(h <= t && i+1-k > q[h]) h++;
        while(h <= t && a[i] >= a[q[t]]) t--;
        
        q[++t] = i;
        if(i >= k-1) printf("%d ",a[q[h]]);
    }
    return 0;
}

我们可以调试一下看一看各个变量的变化情况,这个过程会更加清晰明了

#include 
using namespace std;

const int N = 1e6+10;
int a[N],q[N];

int main()
{
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
    
    int h = 0, t = -1;//h-队头下标, t-队尾下标
    for (int i = 0; i < n; i ++ )
    {
        if(h <= t && i+1-k > q[h]) h++;
        while(h <= t && a[i] <= a[q[t]]) t--;
        
        q[++t] = i;
        // if(i >= k-1) printf("%d ",a[q[h]]);
        
        printf("i=%d, h=%d, t=%dnq[]:", i, h, t);
        for (int j = h; j <= t; j ++ ){
            cout << a[q[j]] << ' ';
        }
        cout << endl;
        
        if(i >= k-1) printf("min=%dn",a[q[h]]);
    }
    
    cout << "==============" << endl;
    h = 0, t = -1;
    for (int i = 0; i < n; i ++ )
    {
        if(h <= t && i+1-k > q[h]) h++;
        while(h <= t && a[i] >= a[q[t]]) t--;
        
        q[++t] = i;
        // if(i >= k-1) printf("%d ",a[q[h]]);
        
        printf("i=%d, h=%d, t=%dnq[]:", i, h, t);
        for (int j = h; j <= t; j ++ ){
            cout << a[q[j]] << ' ';
        }
        cout << endl;
        
        if(i >= k-1) printf("max=%dn",a[q[h]]);
    }
    return 0;
}