题目描述
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.
解题思路
- 如果辅助栈为空，则还需要将元素插入到辅助栈
- 如果不为空，比较插入的元素和辅助栈的top大小，如果小于等于辅助栈的top大小，则还需要插入辅助栈
- 弹出的时候，如果s1弹出的元素等于s2栈的头元素，则还需要将s2弹出
class MinStack {
public:
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
s1.push(x);
if(s2.empty()){
s2.push(x);
}else if(x <= s2.top()){
s2.push(x);
}
}
void pop() {
int x = s1.top();
s1.pop();
if(x == s2.top()){
s2.pop();
}
}
int top() {
return s1.top();
}
int getMin() {
return s2.top();
}
private:
stack<int> s1;
stack<int> s2;
};
/**
* 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();
*/
