155. Min Stack
1. Question
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.
2. Examples
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
3. 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 to
push
,pop
,top
, andgetMin
.
4. References
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/min-stack 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
5. Solutions
class MinStack {
Deque<Integer> input;
Deque<Integer> sorted;
public MinStack() {
input = new LinkedList<>();
sorted = new LinkedList<>();
}
public void push(int val) {
input.push(val);
if (sorted.isEmpty() || sorted.peek() >= val) {
sorted.push(val);
}
}
public void pop() {
if (sorted.peek().equals(input.pop())) {
sorted.pop();
}
}
public int top() {
return input.peek();
}
public int getMin() {
return sorted.peek();
}
}