【23】155. Min Stack-优快云博客

本文介绍了一种特殊栈的设计——MinStack，它支持常规的栈操作，并能在常数时间内检索到栈中的最小元素。通过使用两个栈s1和s2，确保了在进行push、pop、top和getMin等操作时的高效性。

155. Min Stack

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Difficulty: Easy
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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.

Example:

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.

 1 class MinStack {
 2 public:
 3     /** initialize your data structure here. */
 4     MinStack() {
 5         
 6     }
 7     
 8     void push(int x) {
 9         s1.push(x);
10         if(s2.empty() || s2.top() >= x) s2.push(x);//要大于等于！重复值要存！
11     }
12     
13     void pop() {
14         if(s2.top() == s1.top() && !s2.empty()) s2.pop();
15         s1.pop();
16     }
17     
18     int top() {
19         return s1.top();
20     }
21     
22     int getMin() {
23         return s2.top();
24     }
25     
26 private:
27     stack<int> s1, s2;
28 };
29 
30 /**
31  * Your MinStack object will be instantiated and called as such:
32  * MinStack obj = new MinStack();
33  * obj.push(x);
34  * obj.pop();
35  * int param_3 = obj.top();
36  * int param_4 = obj.getMin();
37  */