LC.155. Min Stack(优化，针对整块一样数传入)

如果有特别多重复的数进来的情况，并且是一整块一样的数进来，那是可以被优化的。

1111111  222222  -1-1-1-1 111111 -2-2-2-2-2 33333 

如果 进来的数字是交叉的，则优化空间有限: 1212121212

public class LC_155_MinStack_II {

    private Deque<Integer> stack ;
    private Deque<Integer> minStack ;
    //keep each minStack item's times
    private Deque<Integer> counter ;

    /** initialize your data structure here. */
    public LC_155_MinStack_II() {
        stack = new LinkedList<>() ;
        minStack = new LinkedList<>();
        counter = new LinkedList<>();
    }
    /*
     stack      -2  -2 -2  0  0  -3
     min        -2 -3
     counter    5   1
    * */
    public void push(int x) {
        stack.push(x);
        if (minStack.isEmpty()){
            minStack.push(x);
            counter.push(1); //initialize new item
        } else{
            if (minStack.peek() < x){
                int times = counter.pop();
                counter.push(times+1);
            } else{
                minStack.push(x);
                counter.push(1);
            }
        }
    }
    /*
        stack      -2  -2 -2  0  0  -3
        min        -2  -3
        counter    5   1
       * */
    public void pop() {
        if (stack.isEmpty()) return ;
        stack.pop();
        //handle the min: only pop minStack when the counter's item reach 0
        int times = counter.pop();
        if (times>1){
            times--;
            counter.push(times);
        }
        //when counter == 1 or less(shouldnt possible be <0), then directly pop counter
        else {
            minStack.pop();
        }
    }
    //peek
    public int top() {
        if (stack.isEmpty()) return -1 ;
        return stack.peek() ;
    }

    public int getMin() {
        if (minStack.isEmpty()) return -1 ;
        return minStack.peek();
    }
}