题目描述:
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.
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
题目解释:
设计一个栈包含push、pop、top和返回最小值这几个操作。
题目解法:
1.我的解法。设计一个结构,然后实现栈的功能。代码如下:
class MinStack {
/** initialize your data structure here. */
private Node head;
public MinStack() {
head = new Node(-1);
}
public void push(int x) {
Node node = new Node(x);
node.next = head.next;
head.next = node;
}
public void pop() {
if(head.next != null) {
Node node = head.next;
head.next = node.next;
}
}
public int top() {
return head.next.val;
}
public int getMin() {
Node work = head.next;
int min = work.val;
while(work != null) {
if(work.val < min) min = work.val;
work = work.next;
}
return min;
}
class Node {
int val;
Node next;
public Node(int x) {
this.val = x;
}
}
}
/**
* 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();
*/
2.讨论区的解法,牺牲了一部分的空间换取时间。
class MinStack {
private Node head;
public void push(int x) {
if(head == null)
head = new Node(x, x);
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 getMin() {
return head.min;
}
private class Node {
int val;
int min;
Node next;
private Node(int val, int min) {
this(val, min, null);
}
private Node(int val, int min, Node next) {
this.val = val;
this.min = min;
this.next = next;
}
}
}