python stack example
In Python, you can use list to implement the stack. 1. Use append() to press the stack, 2. Use pop() to pop the stack, 3. View the top of the stack through the index [-1], 4. Use len() to determine whether it is empty, 5. Encapsulating the Stack class can improve the readability of the code. All operations are based on the end to ensure the O(1) time complexity, and are suitable for function calls, bracket matching, DFS, undo operations and other scenarios.
The stack in Python is a "last in first out" (LIFO, Last In First Out) data structure. Although Python does not have a special built-in Stack type, we can easily implement the basic operations of the stack with list : push and pop .
Here is a simple Python stack usage example:
✅ Use list to implement stack
# Create an empty stack = []
# push - use the append() method stack.append(10)
stack.append(20)
stack.append(30)
print("Current stack:", stack) # Output: [10, 20, 30]
#Pop-Use the pop() method (remove and return the last element)
top_item = stack.pop()
print("Popular element:", top_item) # Output: 30
print("Current stack:", stack) # Output: [10, 20]
# View the top element of the stack (not deleted)
if stack:
print("top element:", stack[-1]) # Output: 20
# determine whether the stack is empty is_empty = len(stack) == 0
print("Stack is empty?", is_empty) # Output: False?️ Summary of common stack operations
| operate | method | illustrate |
|---|---|---|
| push(x) | stack.append(x) | Push element x to the top of the stack |
| pop() | stack.pop() | Remove and return the top element of the stack. An error will be reported when the stack is empty. |
| peek/top() | stack[-1] | View the top element of the stack without removing it |
| is_empty() | len(stack) == 0 | Determine whether the stack is empty |
| size() | len(stack) | Get the number of elements in the stack |
⚠️ Note: When using
pop(), make sure that the stack is not empty, otherwiseIndexErrorwill be thrown. You can first determine whether it is empty.
✅ Safer stack operation (with empty check)
def safe_pop(stack):
if not stack:
print("Stack is empty, element cannot be popped")
return None
return stack.pop()
def safe_peek(stack):
if not stack:
print("Stack is empty, no top element")
return None
return stack[-1]
# Example stack = [1, 2, 3]
print(safe_peek(stack)) # Output: 3
print(safe_pop(stack)) # Output: 3
print(safe_pop(stack)) # Output: 2
print(stack) # Output: [1]
print(safe_pop(stack)) # Output: 1
print(safe_pop(stack)) # Output: The stack is empty, element cannot be popped? Optional: Encapsulate a Stack class
If you want the code to be clearer and more object-oriented, you can encapsulate a Stack class:
class Stack:
def __init__(self):
self.items = []
def push(self, item):
self.items.append(item)
def pop(self):
if not self.is_empty():
return self.items.pop()
raise IndexError("pop from empty stack")
def peek(self):
if not self.is_empty():
return self.items[-1]
raise IndexError("peek from empty stack")
def is_empty(self):
return len(self.items) == 0
def size(self):
return len(self.items)
def __str__(self):
return str(self.items)
# Use examples s = Stack()
s.push(1)
s.push(2)
s.push(3)
print("stack:", s) # [1, 2, 3]
print("top:", s.peek()) # 3
print("pop:", s.pop()) # 3
print("Stack Size:", s.size()) # 2? Practical application scenarios
- Function call stack (automatic maintenance of the system)
- Expression evaluation (such as bracket matching)
- Depth First Search (DFS)
- Undo function
- Browser's Back button
Basically that's it. Use list to achieve simple and efficient stack, suitable for most scenarios. As long as you are careful not to insert/delete in the middle, and only at the end of the operation, you can ensure the O(1) time complexity.
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