【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
155. Min Stack
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
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push(x) -- Push element x onto stack.
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pop() -- Removes the element on top of the stack.
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top() -- Get the top element.
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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.
//实现一个最小栈(要求获取最小值用常数时间)
//方法一：用两个栈，一个栈存数据，一个栈存各阶段最小数
class MinStack
{
private:
stack<int> s1; //存数据
stack<int> s2; //存各阶段最小数
public:
/** initialize your data structure here. */
MinStack()
{
}
void push(int x)
{
s1.push(x);
if(s2.empty() || x <= s2.top()) s2.push(x); //如果s2为空，或者存入数小于等于之前最小数，则传入s2
}
void pop()
{
if(s1.top() == s2.top()) s2.pop(); //如果pop的是当前极小值，则s2也跟着pop
s1.pop();
}
int top()
{
return s1.top();
}
int getMin()
{
return s2.top(); //栈顶元素为当前最小数
}
};
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
比较min栈与max队列
（1） if(s2.empty() || x <= s2.top()) s2.push(x); //如果s2为空，或者存入数小于等于之前最小数，则传入s2
（2）
while(!index.empty() && num[i] >= num[index.back()])
index.pop_back(); //从队尾依次弹出队列中比当前num值小的元素，同时也能保证队列首元素为当前窗口最大值下标
index.push_back(i); //插入当前索引值，因为可能为其他窗口下的最大值