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:
MinStack minStack = new MinStack(); minStack.push(-2); minStack.push(0); minStack.push(-3); minStack.getMin(); --> Returns -3. minStack.pop(); minStack.top(); --> Returns 0. minStack.getMin(); --> Returns -2.
Tips：本题主要想实现求栈中最小值。使用两个栈，其中一个为基本栈，另外一个用于保存最小值。
保存最小值的栈的操作思路如下：
push（）：当栈为空，或者即将压入的数字小于minstack的栈顶时，执行压入操作。
pop（）：当要弹出的数字，正好等于minstack栈顶的时候才弹出minStack的栈顶。
getMin（）：直接返回minstack的栈顶即可。
package easy;
import java.util.Stack;
public class L155MinStack {
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> minstack = new Stack<Integer>();
public void push(int x) {
stack.push(x);
System.out.println(" stack push:"+x);
if (minstack.isEmpty() || x <= minstack.peek()) {
minstack.push(x);
System.out.println("minstack push:"+x);
}
}
public void pop() {
int x = stack.peek();
stack.pop();
System.out.println("stack Pop:" + x);
if (x == minstack.peek()) {
minstack.pop();
System.out.println("Minstack Pop:" + x);
}
}
public int top() {
System.out.println("peek" + stack.peek());
return stack.peek();
}
public int getMin() {
System.out.println("minStack peek" + minstack.peek());
return minstack.peek();
}
public static void main(String[] args) {
L155MinStack obj = new L155MinStack();
obj.push(2);
obj.push(1);
obj.push(3);
obj.push(4);
obj.pop();
obj.top();
}
}
