tag:
- Easy;
question:
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.
思路：
本题最小栈跟原来的栈相比就是多了一个功能，可以返回该栈的最小值。使用两个栈来实现，一个栈来按顺序存储push进来的数据，另一个用来存出现过的最小值。代码如下:
class MinStack {
public:
/** initialize your data structure here. */
MinStack() {}
void push(int x) {
s1.push(x);
if (s2.empty() || x <= s2.top()) s2.push(x);
}
void pop() {
if (s1.top() == s2.top()) s2.pop();
s1.pop();
}
int top() {
return s1.top();
}
int getMin() {
return s2.top();
}
private:
stack<int> s1, 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();
*/
