155. Min Stack - Unsolved

时间：2017-06-10 22:33:33 阅读：248 评论：0 收藏：0 [点我收藏+]

标签：amp tac discuss pytho obj object .com trie runtime

https://leetcode.com/problems/min-stack/#/solutions

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.

Sol：

class MinStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []
        # set two stacks
        self.minStack = []

    def push(self, x):
        self.stack.append(x)
        if len(self.minStack) and x == self.minStack[-1][0]
        
        

    def pop(self):
        self.minStack.pop()
        return self.stack.pop()
        

    def top(self):
        """
        # :rtype: int
        """
        return self.stack[-1]
        

    def getMin(self):
        """
        # :rtype: int
        """
        
        return self.minStack[ len(self.minStack) - 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()

Submission Result: Runtime Error More Details 

Runtime Error Message:
Line 44: IndexError: list index out of range
Last executed input:
["MinStack","push","push","push","getMin","top","pop","getMin"]
[[],[-2],[0],[-1],[],[],[],[]]