Set of Stacks - nowcoder


https://blog.csdn.net/HelloZEX/article/details/81126850
请实现一种数据结构SetOfStacks,由多个栈组成,其中每个栈的大小为size,当前一个栈填满时,新建一个栈。该数据结构应支持与普通栈相同的push和pop操作。

给定一个操作序列int[][2] ope(C++为vector<vector<int>>),每个操作的第一个数代表操作类型,若为1,则为push操作,后一个数为应push的数字;若为2,则为pop操作,后一个数无意义。请返回一个int[][](C++为vector<vector<int>>),为完成所有操作后的SetOfStacks,顺序应为从下到上,默认初始的SetOfStacks为空。保证数据合法。

思路:首先建立一个栈组A(定义为vector<vector<int>> A),存放长度达到size的栈。定义vector<int> vect

操作如下:

从i = 0 ~ ope.size() - 1遍历

1、对于压栈指令(即ope[i][0] = 1),首先判断A是否为空

    1)如果A为空,那么考虑将数据压入vect

            1.如果vect长度已经达到size,那么先将vect压入A,然后清空vect里的元素,再将数据ope[i][1]压入vect;

            2.如果vect长度小于size,直接将vect压入A(此时不可以判断vect的长度是否到达size,即使到达size,也要确定下一个指令不是pop,才可以将vect压入A,以免压入之后,再从A[A.size() - 1]中弹出数据)。

    2)如果A不为空,先判断A的顶栈是否是满的,即判断A[A.size() - 1]长度是不是达到size,如果没有达到size,则将数据压入A[A.size() -1];如果达到size,则对vect进行判断,如果vect长度小于size,直接压入vect,否则执行上一步的第一条。

2、对于出栈指令,首先判断vect是否为空

    1)如果vect为空,则从A的顶栈中弹出顶元素,即执行A[a.size() - 1].pop_back(),弹出之后如果A[a.size() - 1]为空,则从A中弹出A[a.size() - 1],即执行A.pop_back;

    2)如果vect不为空,则直接从vect中弹出元素,即执行vect.pop_back().

遍历结束之后,如果vect不为空,则将vect压入A,即A.push_back(vect)。

返回A。

    vector<vector<int> > setOfStacks(vector<vector<int> > ope, int size) {
        // write code here
        vector<int> vect;
        vector<vector<int>> A;
        int len = vect.size();
        for(int i = 0; i < ope.size(); i ++)
            {
            //压入操作时
            if(ope[i][0] == 1)
                {
                //判断A是否为空
                if(A.size() == 0)
                    {
                    //判断vect是否已满
                    if(vect.size() == size)
                        {
                        A.push_back(vect);
                        vect.clear();
                        vect.push_back(ope[i][1]);
                    }
                    else
                        vect.push_back(ope[i][1]);
                }
                else  //A不为空
                    {
                    //判断A的顶栈是否已满
                    if(A[A.size() - 1].size() < size)
                        A[A.size() - 1].push_back(ope[i][1]);
                    else
                        {
                        if(vect.size() == size)
                            {
                            A.push_back(vect);
                            vect.clear();
                            vect.push_back(ope[i][1]);
                        }
                        else
                            vect.push_back(ope[i][1]);
                    }
                }
            }
            else   //弹出操作
                {
                //判断vect是否为空
                if(vect.empty())   //为空
                    {
                    //弹出A顶栈的元素
                    A[A.size() - 1].pop_back();
                    //判断顶栈是否为空
                    if(A[A.size() - 1].empty())  //为空则弹出顶栈
                        A.pop_back();
                }
                else    //vect不为空
                    vect.pop_back();
            }
        }
        if(!vect.empty())
            A.push_back(vect);
        return A;
    }
https://blog.csdn.net/qq_17034717/article/details/51364533
https://www.nowcoder.com/questionTerminal/69f0ffed01c741c5ae5594a23f7cd739
  public ArrayList<ArrayList<Integer>> setOfStacks(int[][] ope, int size) {
    // write code here
    ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();
    ArrayList<Integer> curArray = new ArrayList<Integer>(size);
    list.add(curArray);
    for (int i = 0; i < ope.length; i++) {
      switch (ope[i][0]) {
      // 1:push
      case 1:
        // 当前数组未满
        if (curArray.size() != size) {
          curArray.add(ope[i][1]);
        } else {
          curArray = new ArrayList<Integer>(size);
          list.add(curArray);
          curArray.add(ope[i][1]);
        }
        break;
      // 2:pop
      case 2:
        // 当前数组不为空
        if (curArray.size() != 0) {
          curArray.remove(curArray.size() - 1);
        } else {
          list.remove(list.size() - 1);
          curArray = list.get(list.size() - 1);
          curArray.remove(curArray.size() - 1);
        }
        break;
      }
    }
    return list;

  }

https://www.bbsmax.com/A/D854p0w5Eg/

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