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.
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,必须能够以O(1)的时间复杂度得到栈中的最小值。具体方法是利用两个栈,一个按压栈顺序存储数值,一个按照当前最小值存储数值。
class MinStack {
public:
/** initialize your data structure here. */
MinStack() {}
void push(int x) {
s1.push(x);
if(s2.empty()) s2.push(x);
else s2.push(min(s2.top(),x));
}
void pop() {
s1.pop();
s2.pop();
}
int top() {
return s1.top();
}
int getMin() {
return s2.top();
}
private:
stack<int> s1;
stack<int> s2;
};