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.
1 class MinStack { 2 private Stack<Integer> stack1; 3 private Stack<Integer> stack2; 4 /** initialize your data structure here. */ 5 public MinStack() { 6 this.stack1 = new Stack<Integer>(); 7 this.stack2 = new Stack<Integer>(); 8 } 9 10 public void push(int x) { 11 this.stack1.push(x); 12 if(this.stack2.isEmpty()){ 13 this.stack2.push(x); 14 }else if(x < this.stack2.peek()){ 15 this.stack2.push(x); 16 }else{ 17 this.stack2.push(stack2.peek()); 18 } 19 } 20 21 public void pop() { 22 this.stack1.pop(); 23 this.stack2.pop(); 24 } 25 26 public int top() { 27 return this.stack1.peek(); 28 } 29 30 public int getMin() { 31 return this.stack2.peek(); 32 } 33 } 34 35 /** 36 * Your MinStack object will be instantiated and called as such: 37 * MinStack obj = new MinStack(); 38 * obj.push(x); 39 * obj.pop(); 40 * int param_3 = obj.top(); 41 * int param_4 = obj.getMin(); 42 */
2019-03-03 16:21:27