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, only nodes itself may be 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
public class Solution { public ListNode reverseKGroup(ListNode head, int k) { if(k<1||head == null) return head; ListNode prev = new ListNode(0); prev.next = head; head = prev; ListNode cur = prev.next; while(cur!=null ){ int counter = k; while(cur!=null &&counter >1){ cur = cur.next; counter--; } if(cur!=null){ cur = prev.next;//prev link to cur counter = k; while(counter >1){// greate than 1 counter--; ListNode temp = cur.next; cur.next = temp.next; temp.next = prev.next; prev.next = temp; } prev = cur; cur = prev.next; } } return head.next; } }
mistakes: 条换顺序的时候要挨着，现在等号右边出现，然后左边出现
learned: when swap, swap one by one just like swap number// order in swap does matter!
process: k=3 1 - 2- 3 -4 -5 ----> 2 -> 1 ->3 -> 4 ->5 -------> 3 -> 2 -> 1-> 4 -> 5
public class Solution { public ListNode reverseKGroup(ListNode head, int k) { // Start typing your Java solution below // DO NOT write main() function if (head ==null || k==1){ return head; } ListNode dummy =new ListNode (0); dummy.next=head; ListNode pre=dummy; int i=0; while (head!=null){ i++; if (i%k==0){ pre=reverse(pre,head.next); head=pre.next; } else { head=head.next; } } return dummy.next; } /** * Reverse a link list between pre and next exclusively * an example: * a linked list: * 0->1->2->3->4->5->6 * | | * pre next * after call pre = reverse(pre, next) * * 0->3->2->1->4->5->6 * | | * pre next * @param pre * @param next * @return the reversed list's last node, which is the precedence of parameter next */ public ListNode reverse(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; } }