題目:棧的最小值
具體描述:
請設計一個棧,除了常規棧支持的pop與push函數以外,還支持min函數,該函數返回棧元素中的最小值。執行push、pop和min操作的時間複雜度必須爲O(1)。
示例:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.
來源:力扣(LeetCode)
鏈接:https://leetcode-cn.com/problems/min-stack-lcci
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題解:
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack1=[] #棧
self.stack2=[] #存放最小值的list
def push(self, x: int) -> None:
self.stack1.append(x)
if len(self.stack2) == 0:
self.stack2.append(x)
else:
if(self.stack2[-1] > x):
self.stack2.append(x)
else:
self.stack2.append(self.stack2[-1])
def pop(self) -> None:
self.stack1.pop()
self.stack2.pop()
def top(self) -> int:
return(self.stack1[-1])
def getMin(self) -> int:
return(self.stack2[-1])
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()