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。