Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the?MinStack?class:
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:
- 数据栈+辅助栈
- 数据栈
法一:?
class MinStack {
// Time: O(1)
// Space: O(N)
private Deque<Integer> stack;
private Deque<Integer> minStack;
public MinStack() {
stack = new LinkedList<Integer>();
minStack = new LinkedList<Integer>();
minStack.addFirst(Integer.MAX_VALUE);
}
public void push(int val) {
stack.addFirst(val);
minStack.addFirst(Math.min(minStack.peek(), val));
}
public void pop() {
stack.removeFirst();
minStack.removeFirst();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(val);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
法二:?
class MinStack {
// Time: O(1)
// Space: O(N)
private Deque<Integer> stack;
private int min = Integer.MAX_VALUE;
public MinStack() {
stack = new LinkedList<Integer>();
}
public void push(int val) {
if (val <= min) {
stack.addFirst(min);
min = val;
}
stack.addFirst(val);
}
public void pop() {
if (stack.removeFirst() == min) {
min = stack.removeFirst();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(val);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/