155

Author: tech-cow

Diff for: README.md

@@ -54,3 +54,9 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 ## 4/17 Tasks (Stacks and Queues: CtCi Chapter 3)
 |  #  | Title | Solution | Time | Space | Difficulty |Tag| Note|
 |-----|-------| -------- | ---- | ------|------------|---|-----|
+
+
+
+
+## 4/18 Tasks (Stacks Easy)
+|155|[Min Stack](https://leetcode.com/problems/min-stack/#/description)| [Python](./stack_queue/minStack.py) | _O(1)_| _O(n)_  | Easy | ||

Diff for: stack_queue/minStack.py

@@ -0,0 +1,48 @@
+#!/usr/bin/python
+# -*- coding: utf-8 -*-
+
+# Author: Yu Zhou
+
+# ****************
+# 155. Min Stack
+# Time: O(1)
+# Space: O(n)  using extra stack
+
+# Descrption:
+# 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.
+# ****************
+
+# 思路：
+# 创建一个额外的stack用来储存所有的最小值
+# 返回Min的时候，就返回minstack的最后一个值就好了
+
+
+# ****************
+# Final Solution *
+# ****************
+class MinStack(object):
+
+    def __init__(self):
+        self.stack = []
+        self.minstack = []
+
+    def push(self, x):
+        self.stack.append(x)
+        if self.minstack:
+            x = min(self.minstack[-1], x)
+        self.minstack.append(x)
+
+    def pop(self):
+        self.minstack.pop()
+        self.stack.pop()
+
+    def top(self):
+        return self.stack[-1]
+
+    def getMin(self):
+        return self.minstack[-1]

Diff for: stack_queue/template.py

@@ -0,0 +1,29 @@
+#!/usr/bin/python
+# -*- coding: utf-8 -*-
+
+# Author: Yu Zhou
+
+# ****************
+# Descrption:
+# 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.
+# ****************
+
+# 思路：
+# 创建一个额外的stack用来储存所有的最小值
+# 返回Min的时候，就返回minstack的最后一个值就好了
+
+
+
+# ****************
+# Final Solution *
+# ****************
+
+
+# ***************************************
+# The following code is an fail attempt *
+# ***************************************