剑指offer--面试题30:包含min函数的栈

#include<iostream.h>#include<stack>using namespace std;class Solution{public:    stack<int> st, minSt;//st表示存储元素的栈，minSt保存每次栈操作时，保存栈中最小元素void push(int value) ;void pop() ;int min() ;   };void Solution::push(int value) {st.push(value);if (minSt.empty()||value < minSt.top())minSt.push(value);else minSt.push(minSt.top()); return;}void Solution::pop() {//出栈时，需要对两个栈st和minSt同时操作if (!st.empty()) {st.pop();minSt.pop();}return;}int Solution::min() {if (!minSt.empty()) {return minSt.top();}return 0;}void main(){Solution s;s.push (2);s.push (3);s.push (1);s.push (7);s.push (0);s.push (13);s.pop();s.pop();s.pop();printf("当前最小栈内元素为：%d\n",s.min());}