Stack konteyneri. Stek usulda amallar. Steklarni massiv va ro‘yxatlar yordamida ishlov berish Bajardi: tt301-19 gr

isEmpty() - stek bo'sh bo'lsa True, aks holda False qaytaradi.

peek() - stekdagi eng yuqori elementni qaytaring. Agar stek bo'sh bo'lsa, istisno qiling.

push(item) - qiymatni stek boshiga suring.

pop() - stek boshidagi qiymatni olib tashlang va qaytaring. Agar stek bo'sh bo'lsa, istisno qiling.

Quyida yuqoridagi yondashuvni amalga oshirish ko'rsatilgan:

class Node: def __init__(self, value): self.value = value self.next = None class Stack: def __init__(self): self.head = Node("head") self.size = 0 def __str__(self): cur = self.head.next out = "" while cur: out += str(cur.value) + "->" cur = cur.next return out[:-3] def getSize(self): return self.size def isEmpty(self): return self.size == 0 def peek(self): if self.isEmpty(): raise Exception("Stekka element qo'shish") return self.head.next.value def push(self, value): node = Node(value) node.next = self.head.next self.head.next = node self.size += 1 def pop(self): if self.isEmpty(): raise Exception("Stekdan elementlari olish") remove = self.head.next self.head.next = self.head.next.next self.size -= 1 return remove.value # Driver Code if __name__ == "__main__": stack = Stack() for i in range(1, 11): stack.push(i) print(f"Stek: {stack}") for _ in range(1, 6): remove = stack.pop() print(f"Olindi: {remove}") print(f"Stek elementlari: {stack}")

Output: 
Stek: 10->9->8->7->6->5->4->3->2->
Olindi: 10
Olindi: 9
Olindi: 8
Olindi: 7
Olindi: 6
Stek elementlari: 5->4->3->2->

Xulosa
Men 301-19 guruh talabasi ______________________________ ushbu mustaqil ishni bajarish davomida yarimstatik ma’lumotlar tuzilmasi va navbat tuzilmalarini ishlash algoritmlarini tahlil qilib chiqdim.


Menga berilgan individual topshiriqni algoritm va dasturiy ta’minotini ishlab chiqdim.