LeetCode--155--最小栈(java版)

时间:2022-05-22 11:30:58

设计一个支持 push,pop,top 操作,并能在常数时间内检索到最小元素的栈。

  • push(x) -- 将元素 x 推入栈中。
  • pop() -- 删除栈顶的元素。
  • top() -- 获取栈顶元素。
  • getMin() -- 检索栈中的最小元素。

示例:

MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.
 class MinStack {
private Stack<Integer> stack1;
private Stack<Integer> stack2;
/** initialize your data structure here. */
public MinStack() {
this.stack1 = new Stack<Integer>();
this.stack2 = new Stack<Integer>();
} public void push(int x) {
this.stack1.push(x);
if(this.stack2.isEmpty()){
this.stack2.push(x);
}else if(x < this.stack2.peek()){
this.stack2.push(x);
}else{
this.stack2.push(stack2.peek());
}
} public void pop() {
this.stack1.pop();
this.stack2.pop();
} public int top() {
return this.stack1.peek();
} public int getMin() {
return this.stack2.peek();
}
} /**
* 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();
*/

2019-03-03 16:21:27