201705031-leetcode-155-Min Stack

1.Description

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.

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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.

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解讀
題目中要常數時間內能夠解決該問題，所以應當考慮把最小值同時壓棧的操作

2.Solution

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.stack = []

    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        #這裏push數據的時候，把最小值也同時壓入，已tuple的方式壓入
        min = self.getMin()
        if min == None or x < min:
            min = x
        self.stack.append((x, min))

    def pop(self):
        """
        :rtype: void
        """
        if self.stack == []:
            return None
        self.stack.pop()

    def top(self):
        """
        :rtype: int
        """
        if self.stack == []:
            return None
        return self.stack[-1][0]

    def getMin(self):
        """
        :rtype: int
        """
        if self.stack == []:
            return None
        return self.stack[-1][1]