Implement a stack with min() function, which will return the smallest number in the stack.

It should support push, pop and min operation all in O(1) cost.

 Notice

min operation will never be called if there is no number in the stack.

来源： 

分析

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

public  class  MinStack {      private  Stack<Integer> stack =  new  Stack<Integer>();      private  Stack<Integer> stack_min =  new  Stack<Integer>();            public  MinStack() {          // do initialize if necessary      }            public  void  push( int  number) {          // write your code here          stack.push(number);          if (!stack_min.empty() && stack_min.peek() < number){              stack_min.push(stack_min.peek());          }          else {              stack_min.push(number);          }      }        public  int  pop() {          // write your code here          int  top = - 1 ;          if (!stack.empty()){              top = stack.peek();              stack.pop();              stack_min.pop();          }          return  top;      }        public  int  min() {          // write your code here          return  stack_min.empty() ? - 1  : stack_min.peek();      } }