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.
这道题看上去简单，但是注意在出栈的时候，如果min的出去了，怎么破？
class MinStack {
Stack<Integer> s = new Stack<Integer>();
Stack<Integer> ms = new Stack<Integer>();
int min = Integer.MAX_VALUE;
/** initialize your data structure here. */
public MinStack() {
}
public void push(int x) {
s.push(x);
if(min>x){
min = x;
}
ms.push(min);
}
public void pop() {
s.pop();
ms.pop();
if(s.isEmpty()){
min = Integer.MAX_VALUE;
}
else{
min = ms.peek();
}
}
public int top() {
return s.peek();
}
public int getMin() {
return min;
}
}
/**
* 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();
*/
微信扫一扫
支付宝扫一扫
