leetcode406

public class Solution {
    public int[,] ReconstructQueue(int[,] people) {
        if (people == null || people.Length == 0)
            {
                return new int[,] { };
            }

            var row = people.GetLength(0);//二元组个数
            var col = people.GetLength(1);//2

            var dic = new Dictionary<int, List<int>>();

            var ary = new int[row, col];

            //将前面为0的“队头”确定
            for (int i = 0; i < row; i++)
            {
                var height = people[i, 0];
                var position = people[i, 1];

                if (!dic.ContainsKey(position))
                {
                    var po = new List<int>();
                    po.Add(height);
                    dic.Add(position, po);
                }
                else
                {
                    dic[position].Add(height);
                }
            }

            //先确定队头
            var headlist = dic[0].OrderBy(x => x).ToList();
            for (int i = 0; i < headlist.Count; i++)
            {
                ary[i, 0] = headlist[i];
                ary[i, 1] = 0;
            }
            //按照positon进行插入排序
            var plist = dic.Keys.OrderBy(x => x).ToList();

            var dtcount = dic[0].Count;//队头的二元组数量

            foreach (var p in plist)
            {
                if (p == 0)
                {
                    continue;
                }
                var addlist = dic[p].OrderBy(x => x).ToList();

                for (int i = 0; i < addlist.Count; i++)//循环剩余的列表
                {
                    var curheight = addlist[i];
                    var curposition = p;

                    var cf = 0;//队头中满足条件的数量
                    var inserted = false;//是否已经插入
                    for (int j = 0; j < dtcount; j++)//循环队头，找到第一个不满足的位置
                    {
                        if (curheight <= ary[j, 0])
                        {
                            cf++;//发现一个，比当前元素相等或更高的元素
                            if (cf > p)
                            {
                                //找到了不满足的情况，当前的j为插入的位置

                                //j以及j之后的元素都向后移动
                                for (int k = dtcount - 1; k >= j; k--)
                                {
                                    ary[k + 1, 0] = ary[k, 0];
                                    ary[k + 1, 1] = ary[k, 1];
                                }
                                ary[j, 0] = curheight;
                                ary[j, 1] = curposition;

                                inserted = true;
                                dtcount++;
                                break;
                            }
                        }
                    }

                    if (!inserted)//没有遇到冲突的情况，插入末尾
                    {
                        ary[dtcount, 0] = curheight;
                        ary[dtcount, 1] = curposition;
                        dtcount++;
                    }

                }

            }
            return ary;
    }
}