描述
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack class:
- MinStack() initializes the stack object.
- void push(val) pushes the element val onto the stack.
- void pop() removes the element on the top of the stack.
- int top() gets the top element of the stack.
- int getMin() retrieves the minimum element in the stack.
Example 1:
Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
Output
[null,null,null,null,-3,null,0,-2]
Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top(); // return 0
minStack.getMin(); // return -2
Note:
-2^31 <= val <= 2^31 - 1
Methods pop, top and getMin operations will always be called on non-empty stacks.
At most 3 * 10^4 calls will be made to push, pop, top, and getMin.
解析
根据题意，就是实现各种操作的函数，基本都是对列表的操作，思路简单，不懂的重温 python 语法吧。
解答
class MinStack(object):
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
def push(self, val):
"""
:type val: int
:rtype: None
"""
self.stack.append(val)
def pop(self):
"""
:rtype: None
"""
if self.stack:
self.stack.pop(-1)
def top(self):
"""
:rtype: int
"""
if self.stack:
return self.stack[-1]
return None
def getMin(self):
"""
:rtype: int
"""
if self.stack:
return min(self.stack)
return None
运行结果
Runtime: 616 ms, faster than 25.12% of Python online submissions for Min Stack.
Memory Usage: 16.9 MB, less than 89.88% of Python online submissions for Min Stack.
原题链接：leetcode.com/problems/mi…
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