定义栈的数据结构,请在该类型中实现一个能够得到栈的最小元素的 min 函数在该栈中,调用 min、push 及 pop 的时间复杂度都是 O(1)。
示例:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.min(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.min(); --> 返回 -2.
提示:
各函数的调用总次数不超过 20000 次
题解:
class MinStack {
private Node head;
/** initialize your data structure here. */
public MinStack() {
}
public void push(int x) {
if(head == null) {
head = new Node(x, x, null);
}
else {
head = new Node(x, Math.min(x, head.min), head);
}
}
public void pop() {
head = head.next;
}
public int top() {
return head.val;
}
public int min() {
return head.min;
}
private class Node {
int val;
int min;
Node next;
public Node(int val, int min, Node next) {
this.val = val;
this.min = min;
this.next = next;
}
}
}
//双栈法
// class MinStack {
// Stack<Integer> s,sup;
// /** initialize your data structure here. */
// public MinStack() {
// s=new Stack<Integer>();
// sup=new Stack<Integer>();
// }
// public void push(int x) {
// s.push(x);
// if(sup.empty()||sup.peek()>x)sup.push(x);
// else sup.push(sup.peek());
// }
// public void pop() {
// sup.pop();
// s.pop();
// }
// public int top() {
// return s.peek();
// }
// public int min() {
// return sup.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.min();
*/