剑指offer_20:包含min函数的栈
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
1、用栈做
class MinStack {
Stack<Integer> stack;
Stack<Integer> minStack;
/** initialize your data structure here. */
public MinStack() {
stack=new Stack<>();
minStack=new Stack<>();
}
public void push(int x) {
stack.push(x);
if(minStack.isEmpty()){
minStack.push(x);
}else{
if(minStack.peek()>=x){
//这里用>=而不是>,是因为
//若stack压入两个2,min栈压入一个2;
//再弹出一个2,min栈也弹出一个2,这时min栈为空,但stack栈仍然有最小值
//所以要加上一个=号
minStack.push(x);
}
}
}
public void pop() {
if(stack.peek().equals(minStack.peek())){
minStack.pop();
}
stack.pop();
}
public int top() {
return stack.peek();
}
public int min() {
return minStack.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();
*/
2、用链表做
class MinStack {
private class Node{
int val;
int min;
Node next;
public Node(int val,int min){
this.val=val;
this.min=min;
}
}
private Node head;
/** initialize your data structure here. */
public MinStack() {
}
public void push(int x) {
if(head==null){
head=new Node(x,x);
}else{
//头插法(伪)
Node node=new Node(x,head.min>=x?x:head.min);
node.next=head;
head=node;
}
}
public void pop() {
head=head.next;
}
public int top() {
return head.val;
}
public int min() {
return head.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.min();
*/