设计一个支持 push ,pop ,top 操作,并能在常数时间内检索到最小元素的栈。
实现 MinStack 类:
MinStack()初始化堆栈对象。void push(int val)将元素val推入堆栈。void pop()删除堆栈顶部的元素。int top()获取堆栈顶部的元素。int getMin()获取堆栈中的最小元素。
示例 1:
输入:
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]
输出:
[null,null,null,null,-3,null,0,-2]
解释:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.
提示:
-231 <= val <= 231 - 1pop、top和getMin操作总是在 非空栈 上调用push,pop,top, andgetMin最多被调用3 * 104次
解题答案
class MinStack {
private Node head;
public MinStack() {
}
public void push(int val) {
int min = this.head == null ? val : Math.min(val, this.head.min);
this.head = new Node(val, min, this.head);
}
public void pop() {
this.head = this.head.next;
}
public int top() {
return this.head.val;
}
public int getMin() {
return this.head.min;
}
static class Node {
int min;
int val;
Node next;
private Node(int val, int min, Node next) {
this.min = min;
this.val = val;
this.next = next;
}
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(val);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/