leetcode406

public class Solution {

    public int[,] ReconstructQueue(int[,] people) {

        if (people == null || people.Length == )

            {

                return new int[,] { };

            }

            var row = people.GetLength();//二元组个数

            var col = people.GetLength();//

            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 = ; i < row; i++)

            {

                var height = people[i, ];

                var position = people[i, ];

                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[].OrderBy(x => x).ToList();

            for (int i = ; i < headlist.Count; i++)

            {

                ary[i, ] = headlist[i];

                ary[i, ] = ;

            }

            //按照positon进行插入排序

            var plist = dic.Keys.OrderBy(x => x).ToList();

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

            foreach (var p in plist)

            {

                if (p == )

                {

                    continue;

                }

                var addlist = dic[p].OrderBy(x => x).ToList();

                for (int i = ; i < addlist.Count; i++)//循环剩余的列表

                {

                    var curheight = addlist[i];

                    var curposition = p;

                    var cf = ;//队头中满足条件的数量

                    var inserted = false;//是否已经插入

                    for (int j = ; j < dtcount; j++)//循环队头，找到第一个不满足的位置

                    {

                        if (curheight <= ary[j, ])

                        {

                            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 - ; k >= j; k--)

                                {

                                    ary[k + , ] = ary[k, ];

                                    ary[k + , ] = ary[k, ];

                                }

                                ary[j, ] = curheight;

                                ary[j, ] = curposition;

                                inserted = true;

                                dtcount++;

                                break;

                            }

                        }

                    }

                    if (!inserted)//没有遇到冲突的情况，插入末尾

                    {

                        ary[dtcount, ] = curheight;

                        ary[dtcount, ] = curposition;

                        dtcount++;

                    }

                }

            }

            return ary;

    }

}

https://leetcode.com/problems/queue-reconstruction-by-height/#/description

上面这个写的够长的了，用python，4行就可以实现：

 class Solution:

     def reconstructQueue(self, people):

         res = []

         for i in sorted(people, key = lambda x: (-x[0],x[1])):

             res.insert(i[1], i)

         return res

先按照第一个元素倒序排，再按照第二个元素正序排，然后用insert方法，在指定的index上插入。