[LeetCode] Min Stack

最新推荐文章于 2025-09-05 14:41:09 发布

转载最新推荐文章于 2025-09-05 14:41:09 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4557627.html

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#开发工具

本文介绍了一种使用两个栈来高效实现带有获取最小元素操作的数据结构的方法。其中一个栈存储所有元素，另一个栈则记录遇到的最小元素。通过这种方式，在进行插入、删除操作时能够快速维护最小值。

The idea is to use two stacks.

For push, the first stack records the pushed elements and the second stack records all the minimum (including duplicates) that have been seen.

For pop, we need to compare with the top elements of the two stacks. If they are equal (the top element is also minimum, popping it out will change the minimum), pop both; otherwise, only pop the first stack.

The remaining top and getMin will be trivial: just return the top element of the first and second stack respectively.

The code is as follows.

 1 class MinStack {
 2 public:
 3     void push(int x) {
 4         if (stk1.empty() || x <= stk2.top())
 5             stk2.push(x);
 6         stk1.push(x);
 7     }
 8 
 9     void pop() {
10         if (stk1.top() == stk2.top())
11             stk2.pop();
12         stk1.pop();
13     }
14 
15     int top() {
16         return stk1.top();
17     }
18 
19     int getMin() {
20         return stk2.top();
21     }
22 private:
23     stack<int> stk1; 
24     stack<int> stk2;
25 };