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Copy pathO(1)minStack.cpp
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Copy pathO(1)minStack.cpp
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74 lines (63 loc) · 1.39 KB
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// 实现一个特殊的栈，在实现栈的基本功能的基础上，再返回栈中最小元素的操作
// 要求 1 pop push getMin操作的时间复杂度都是O(1)
// 设计的栈类型可以使用现成的栈结构(栈的实现结构：双向链表结构 / 数组)
/**
* 思想:建立两个栈 1数据栈 2最小栈，每当数据栈有数据压入时将该数据与最小栈栈顶元素进行比较，将较小的值同步压入最小栈
* 弹出时两个栈也同步弹出
*/
#include <stack>
#include <iostream>
#include <algorithm>
using namespace std;
class MyStack{
private:
stack<int> dataStack;
stack<int> minStack;
public:
MyStack(){
}
void push(int obj){
dataStack.push(obj);
if (minStack.empty()){
minStack.push(obj);
}
else {
minStack.push(min(this->getMin(), obj));
}
}
void pop(){
minStack.pop();
dataStack.pop();
}
int top() {
return dataStack.top();
}
int getMin(){
if (!minStack.empty()){
return minStack.top();
}
else{
cerr << "your stack is empty";
}
}
int getSize(){
return dataStack.size();
}
};
int main(){
MyStack s1;
s1.push(1);
s1.push(3);
s1.push(6);
cout <<"top is :"<< s1.top() << endl;
for (int i = 0; i < s1.getSize(); i++){
s1.pop();
}
for (int i = 0; i < 15; i++){
s1.push(i % 5);
}
cout << "top is :" << s1.top() << endl;
cout << "min is:"<<s1.getMin() << endl;
system("pause");
return 0;
}