LeetCode OJ:Min Stack(最小栈问题)

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.

设计一个最小栈,使得返回栈中的最小数的这个操作的时间复杂度为常数,用双栈实现,一个栈作为最小数的保存栈,代码如下:

 1 class MinStack {
 2 public:
 3     void push(int x) 
 4     {
 5         s.push(x);
 6         if(sMin.empty() || x <= sMin.top())
 7             sMin.push(x);
 8     }
 9 
10     void pop() 
11     {
12         if(s.top() == sMin.top()){
13             sMin.pop();
14         }
15         s.pop();
16     }
17 
18     int top() 
19     {
20         return s.top();
21     }
22 
23     int getMin() 
24     {
25         return sMin.top();
26     }
27 private:
28     stack<int> s;
29     stack<int> sMin;
30 };
原文地址:https://www.cnblogs.com/-wang-cheng/p/5014376.html