力扣之最小栈——155

问题

解答

辅助栈

min_stack作为储存最小元素的辅助栈使getMin()函数时间复杂度达到O(1)

class MinStack:
    def __init__(self):
        self.stack = []
        self.min_stack = []
    def push(self, x: int) -> None:
        self.stack.append(x)
        #最小栈无元素，或者有更小的元素插入
        if not self.min_stack or x <= self.min_stack[-1]: 
            self.min_stack.append(x)
    def pop(self) -> None:
        #self.stack.pop()已经实现栈顶元素删除
        if self.stack.pop() == self.min_stack[-1]:
            self.min_stack.pop()
    def top(self) -> int:
        return self.stack[-1]
    def getMin(self) -> int:
        return self.min_stack[-1]

调试结果：