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:
AC
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]: self.minStack[-1] = (x, self.minStack[-1][1] + 1) elif len(self.minStack) == 0 or x < self.minStack[-1][0]: self.minStack.append((x, 1)) def pop(self): #如果 栈顶值 == 最小值栈顶值 if self.top() == self.getMin(): #如果 最小值栈顶元素次数 > 1 if self.minStack[-1][1] > 1: #最小值栈顶元素次数 - 1 self.minStack[-1] = (self.minStack[-1][0], self.minStack[-1][1] - 1) else: #最小值栈顶元素弹出 self.minStack.pop() return self.stack.pop() def top(self): return self.stack[-1] def getMin(self): return self.minStack[-1][0] # 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()
Back - up knowledge:
Implementation of Stack
Some sols:
1
https://discuss.leetcode.com/topic/11985/my-python-solution
2
https://discuss.leetcode.com/topic/37294/python-one-stack-solution-without-linklist
3
http://bookshadow.com/weblog/2014/11/10/leetcode-min-stack/
4
http://www.aichengxu.com/data/720464.htm
5