Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10?5??) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
using System; using System.Collections.Generic; struct Point { public string X; public int Y; public string Z; public Point(string v1, int v2, string v3) : this() { this.X = v1; this.Y = v2; this.Z = v3; } } struct Point2 { public string X; public int Y; public int Z; public Point2(string v1, int v2, int v3) : this() { this.X = v1; this.Y = v2; this.Z = v3; } } class T { static void Main(string[] args) { List<Point> pList = new List<Point>(); List<Point> newList = new List<Point>(); var v = Console.ReadLine().Split(new[] { " " }, StringSplitOptions.RemoveEmptyEntries); Point2 P = new Point2(v[0], int.Parse(v[1]), int.Parse(v[2])); for (int i = 0; i < P.Y; i++) { var temp = Console.ReadLine().Split(new[] { " " }, StringSplitOptions.RemoveEmptyEntries); pList.Add(new Point(temp[0], int.Parse(temp[1]), temp[2])); } var addr = P.X; while (addr != "-1") { var p = pList.FindIndex(x => (x.X == addr)); if (p>=0) { newList.Add(pList[p]); addr = pList[p].Z; } pList.RemoveAt(p); } List<Point> rList = new List<Point>(); var max = newList.Count / P.Z; var c = 0; for (int i = 0; i < max; i++) { for (int j = P.Z - 1; j >= 0; j--) { c = P.Z * i + j; if (c < newList.Count) { rList.Add(newList[c]); } } } int m = max * P.Z; while (m < newList.Count) { rList.Add(newList[m++]); } List<Point> rList2 = new List<Point>(); for (int i = 0; i < rList.Count - 1; i++) { Point tt = new Point(); tt = rList[i]; tt.Z = rList[i + 1].X; rList2.Add(tt); } Point t = new Point(); t = rList[rList.Count - 1]; t.Z = "-1"; rList2.Add(t); foreach (var item in rList2) { Console.WriteLine(item.X + " " + item.Y + " " + item.Z); } } }
02-线性结构3 Reversing Linked List （25 分）
原文：https://www.cnblogs.com/interim/p/9723193.html