Java-链表的部分翻转

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package 链表_递归_栈;import java.util.Random;/* * 问题：给定一个链表，翻转该链表从m到n的位置。要求：直接翻转而非申请新空间 * 以第四个和第八个结点为范围：翻转前的链表：13,96,31,66,71,16,43,78,71,1,翻转后的链表：13,96,31,78,43,16,71,66,71,1, * */public class 链表的部分翻转 {public static void main(String[] args) throws Exception {ListNode h1 = getList(10);System.out.print("翻转前的链表：");ListNode l1 = h1.nextNode;while(l1 != null){System.out.print(l1.value + ",");l1 = l1.nextNode;}System.out.println();System.out.print("翻转后的链表：");ListNode headList = reverse(h1, 4, 8);ListNode l2 = headList.nextNode;while(l2 != null){System.out.print(l2.value + ",");l2 = l2.nextNode;}}//翻转链表public static ListNode reverse(ListNode head, int m, int n) throws Exception{if(m > n){throw new  Exception("输入的翻转的范围有误!");}ListNode Head = head;ListNode cur = head.nextNode;int i;for(i = 0; i < m - 1; i++){//空循环找到m-1处//System.out.println(cur.value + ".");head = cur;cur =  cur.nextNode;}ListNode pre = cur;//pre指向第m-1个结点的位置cur = cur.nextNode ;//第m个结点ListNode pnext;for(; i < n - 1; i++){//结合链表图来分析pnext = cur.nextNode;cur.nextNode = head.nextNode;head.nextNode = cur;pre.nextNode = pnext;cur = pnext;}return Head;}//随机生成一个链表，其中参数length是链表结点的个数public static ListNode getList(int length){Random r = new Random();if(length < 0){return null;}ListNode head = new ListNode();ListNode point = head;if(length == 0){return head;}for(int i = 0; i < length; i++){ListNode p = new ListNode(r.nextInt(100));point.nextNode = p;point = point.nextNode;//System.out.println(point.value + ",");}return head;}}
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