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 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:
- Methods
pop
,top
andgetMin
operations will always be called on non-empty stacks.
pop两次:pop是会动的。min == stack.pop()出去了,所以min要更新为当前的值,再来一次min=stack.pop();
class MinStack { int min = Integer.MAX_VALUE; Stack<Integer> stack = new Stack<Integer>(); public void push(int x) { //push一个以前的min作为垫背 // only push the old minimum value when the current // minimum value changes after pushing the new value x if(x <= min){ stack.push(min); min=x; } stack.push(x); } public void pop() { // if pop operation could result in the changing of the current minimum value, // pop twice and change the current minimum value to the last minimum value. if(stack.pop() == min) min=stack.pop(); } public int top() { return stack.peek(); } public int getMin() { return min; } }