Chain List Flip Series

Problem I

Flip Linked list

Program (compact version)

Non-recursive

ListNode reverselist (ListNode head) {ListNode pre = NULL; ListNode cur = head;while (cur! = null) {ListNode next = Cur.next;cur.next = Pre;pre = Cur;cur = Next;} return pre;}

Problem Ii:swap Nodes in Pairs

Given a linked list, swap every, adjacent nodes and return its head.

For example,
Given 1->2->3->4 , you should return the list as 2->1->4->3 .

Your algorithm should use only constant space. Modify the values in the list, only nodes itself can be changed.

Swap a pair of nodes in the linked list in turn.

Program

public class Solution {public    ListNode swappairs (ListNode head) {ListNode dummy = new ListNode (-1); ListNode pre = head; ListNode connect = Dummy;while (pre! = null && Pre.next! = null) {ListNode next = Pre.next; ListNode cross = Next.next;next.next = Pre;pre.next = Cross;connect.next = Next;connect = Pre;pre = Cross;} return dummy.next = = null? Head:dummy.next;}}

Problem Iii:reverse Nodes in K-group

Given A linked list, reverse the nodes of a linked list K at a time and return its modified list.

If the number of nodes is not a multiple of K then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, and only nodes itself is changed.

Only constant memory is allowed.

For example,
Given This linked list:1->2->3->4->5

For k = 2, you should return:2->1->4->3->5

For k = 3, you should return:3->2->1->4->5

Program

public class Solution {public ListNode Reversekgroup (ListNode head, int k) {if (head = = NULL | | head.next = = N ull | |        K <= 1) {return head;        } ListNode dummy = new ListNode (-1);        Dummy.next = head;        ListNode pre = dummy;        ListNode cur = head;                int cnt = 0;            while (cur! = null) {++cnt;                if (cnt% k = = 0) {//reverse group ListNode last = Reversegroup (pre, cur.next);                Pre = last;            cur = last.next;            } else {cur = cur.next;    }} return dummy.next;        } private ListNode Reversegroup (ListNode Pre, ListNode next) {ListNode last = Pre.next;                ListNode cur = last.next;            while (cur! = next) {Last.next = Cur.next;            Cur.next = Pre.next;            Pre.next = cur;        cur = last.next; } return LAst }}

The difficulty is that each time a node is flipped between two nodes.

Problem Iv:reverse LinkedList II

Reverse a linked list from position m to N. Do it in-place and in One-pass.

For example:
Given 1->2->3->4->5->NULL , m = 2 and n = 4,

Return 1->4->3->2->5->NULL .

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤length of list.

Program

public class Solution {public    ListNode Reversebetween (listnode head, int m, int n) {        if (head = = NULL | | head.nex T = = null) {            return head;        }        ListNode dummy = new ListNode (0);        Dummy.next = head;        ListNode pre = dummy;                while (--m > 0) {            pre = Pre.next;        }                ListNode next = head;        while (--n > 0) {            next = Next.next;        }                Reversebetween (pre, next.next);        return dummy.next;    }        private void Reversebetween (ListNode Pre, ListNode next) {        ListNode last = Pre.next;        ListNode cur = last.next;                while (cur! = next) {            last.next = Cur.next;            Cur.next = Pre.next;            Pre.next = cur;            cur = last.next;}}    }

Find the right location, apply the method of question 3 ~

Chain List Flip Series