题目描述:
方法一(ArrayList):
代码:
class MinStack {
ArrayList<Integer> list;
/** initialize your data structure here. */
public MinStack() {
list = new ArrayList<Integer>();
}
public void push(int x) {
list.add(x);
}
public void pop() {
list.remove(list.size()-1);
}
public int top() {
return list.get(list.size()-1);
}
public int getMin() {
int min = Integer.MAX_VALUE;;
for(Integer i : list){
if(i<min){
min = i;
}
}
return min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
结果:
方法二(Deque):
使用这种方法需要注意:消除栈每pop掉一个栈顶对最小数的影响,所以建立一个mindq用来存储每push一个元素之后产生的最小数,从而deque每pop掉一个数后,mindq也会pop掉经由deque pop的元素产生的最小值,mindq的栈顶依旧是当前deque的最小数。
class MinStack {
Deque<Integer> deque;
Deque<Integer> mindq;
/** initialize your data structure here. */
public MinStack() {
deque = new LinkedList<Integer>();
mindq = new LinkedList<Integer>();
mindq.push(Integer.MAX_VALUE);
}
public void push(int x) {
deque.push(x);
mindq.push(Math.min(mindq.peek(),x));
}
public void pop() {
deque.pop();
mindq.pop();
}
public int top() {
return deque.peek();
}
public int getMin() {
return mindq.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
结果: