155. Min Stack 求栈的最小值

Posted 2020-09-30 Long Long Journey

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

Example:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin();   --> Returns -3.
minStack.pop();
minStack.top();      --> Returns 0.
minStack.getMin();   --> Returns -2.

class MinStack(object):
    dataList = []
    def __init__(self):
        """
        initialize your data structure here.
        """
        self.dataList = []
        
    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        curMin = self.getMin()
        if curMin is None or x < curMin:
            curMin = x
        self.dataList.append([x,curMin])
        
    def pop(self):
        """
        :rtype: void
        """
        self.dataList.pop()
        if len(self.dataList) is 0:
            self.curMin = None
        
    def top(self):
        """
        :rtype: int
        """
        if len(self.dataList) is 0:
            return None
        return self.dataList[-1][0]
        
    def getMin(self):
        """
        :rtype: int
        """
        if len(self.dataList) is 0:
            return None
        return self.dataList[-1][1]
        
# Your MinStack object will be instantiated and called as such:
s = MinStack();
s.push(0);
s.push(1);
s.push(0);
s.getMin();
s.pop();
print(s.dataList)
m = s.getMin();
print(m)

来自为知笔记(Wiz)