動機
複習minstk
Problem
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 elementval
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]ExplanationMinStack minStack = new MinStack();minStack.push(-2);minStack.push(0);minStack.push(-3);minStack.getMin(); // return -3minStack.pop();minStack.top(); // return 0minStack.getMin(); // return -2
Constraints:
-231 <= val <= 231 - 1
- Methods
pop
,top
andgetMin
operations will always be called on non-empty stacks. - At most
3 * 104
calls will be made topush
,pop
,top
, andgetMin
.
Sol
stk同時放目前最小值
class MinStack:
def __init__(self):
self.stk = []
def push(self, val: int) -> None:
if not self.stk:
self.stk.append([val,val])
else:
self.stk.append([val, val if val < self.getMin() else self.getMin()])
def pop(self) -> None:
self.stk.pop()
def top(self) -> int:
return self.stk[-1][0]
def getMin(self) -> int:
return self.stk[-1][1]