LeetCode: Min Stack 解题报告

Min Stack

My Submissions

Question

Solution

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.

Show Tags

Have you met this question in a real interview? Yes No

Discuss

SOLUTION 1:

比较直观。用一个min stack专门存放最小值，如果有比它小 或是相等的（有多个平行的最小值都要单独存放，否则pop后会出问题），

则存放其到minstack.具体看代码：

 1 class MinStack {
 2     Stack<Integer> elements = new Stack<Integer>();
 3     Stack<Integer> minStack = new Stack<Integer>();
 4
 5     public void push(int x) {
 6         elements.push(x);
 7         if (minStack.isEmpty() || x <= minStack.peek()) {
 8             minStack.push(x);
 9         }
10     }
11
12     public void pop() {
13         if (elements.isEmpty()) {
14             return;
15         }
16
17         // 这个地方太蛋疼了，居然要用equals...
18         if (elements.peek().equals(minStack.peek())) {
19             minStack.pop();
20         }
21         elements.pop();
22     }
23
24     public int top() {
25         return elements.peek();
26     }
27
28     public int getMin() {
29         return minStack.peek();
30     }
31 }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/stack/MinStack.java