155. Min Stack
Medium
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(int 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.
You must implement a solution with
O(1)
time complexity for each function.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
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
.
type MinStack struct {
min []int
stack []int
}
func Constructor() MinStack {
return MinStack{ }
}
func (this *MinStack) Push(val int) {
if len(this.stack) == 0 {
this.stack = []int{ val }
this.min = []int{ val }
} else {
this.stack = append(this.stack, val)
if val < this.min[len(this.min) - 1]{
this.min = append(this.min, val)
} else {
this.min = append(this.min, this.min[len(this.min) - 1])
}
}
}
func (this *MinStack) Pop() {
this.stack = this.stack[:len(this.stack) - 1]
this.min = this.min[:len(this.min) - 1]
}
func (this *MinStack) Top() int {
return this.stack[len(this.stack) - 1]
}
func (this *MinStack) GetMin() int {
return this.min[len(this.min) - 1]
}
/**
* Your MinStack object will be instantiated and called as such:
* obj := Constructor();
* obj.Push(val);
* obj.Pop();
* param_3 := obj.Top();
* param_4 := obj.GetMin();
*/