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.
原题链接：https://oj.leetcode.com/problems/min-stack/
取出栈中的最小元素须要常数时间。
能够用两个栈来实现，当中一个栈存储的是所有的元素，而还有一个栈存储的是前一个栈中的最小元素，这样每次更新后一个栈就可以。
public class MinStack {
private List<Integer> stack = new ArrayList<Integer>();
private List<Integer> minstack = new ArrayList<Integer>();
public void push(int x) {
stack.add(x);
if(minstack.isEmpty() || minstack.get(minstack.size()-1) >= x)
minstack.add(x);
}
public void pop() {
if(stack.isEmpty())
return;
int tmp = stack.remove(stack.size()-1);
if(!minstack.isEmpty() && tmp == minstack.get(minstack.size()-1))
minstack.remove(minstack.size()-1);
}
public int top() {
if(!stack.isEmpty())
return stack.get(stack.size()-1);
return -1;
}
public int getMin() {
if(!minstack.isEmpty())
return minstack.get(minstack.size()-1);
return -1;
}
}
