LeetCode_25---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, 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

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翻译：

Code：

package From21;import From21.LeetCode24.ListNode;/** * @author MohnSnow * @time 2015年6月10日 上午11:39:18 *  */public class LeetCode25 {/** * @param argsmengdx *            -fnst */static class ListNode {intval;ListNodenext;ListNode(int x) {val = x;}@Overridepublic String toString() {if (this.next != null) {return val + "---" + this.next.toString();} else {return val + "";}}}//https://leetcode.com/discuss/31921/java-o-n-time-o-1-space-solution//352msApublic static ListNode reverseKGroup(ListNode head, int k) {ListNode dummy = new ListNode(0);dummy.next = head;ListNode temp = dummy;ListNode current = head;ListNode reverse = null;ListNode reverseHead = temp;ListNode reverseTail = temp.next;while (current != null) {reverseHead = temp;reverseTail = temp.next;temp = temp.next;for (int i = 0; i < k; i++) {if (current == null)return dummy.next;current = current.next;}current = temp;for (int i = 0; i < k; i++) {current = current.next;temp.next = reverse;reverse = temp;temp = current;}reverseHead.next = reverse;reverseTail.next = current;reverse = null;temp = reverseTail;}return dummy.next;}//自己写的指针好乱好多好杂啊public static ListNode reverseKGroup1(ListNode head, int k) {ListNode dummy = new ListNode(0);dummy.next = head;ListNode temp = dummy;ListNode current = head;ListNode T2 = null;ListNode reverse = null;ListNode T1 = dummy;while (current != null) {head = current;for (int i = 0; i < k - 1; i++) {if (current == null) {return dummy.next;}current = current.next;}reverse = current;T1.next = reverse;reverse.next = null;T1 = head;current = current.next;head.next = current;temp = head;for (int j = 1; j < k; j++) {temp = head.next;if (temp.next == null) {temp.next = head;break;} else {T2 = temp.next;}temp.next = head;head = temp;temp = T2;}}return dummy.next;}//340msApublic static ListNode reverseKGroup2(ListNode head, int k) {if (head == null) {return head;}ListNode tail = head;for (int i = 1; i < k; i++) {tail = tail.next;if (tail == null) {return head;}}ListNode nextRoundHead = tail.next;//下一轮的第一个数ListNode second = head.next;ListNode first = head;for (int i = 1; i < k; i++) {//reserve the k node groupListNode newSecond = second.next;second.next = first;first = second;second = newSecond;}head.next = reverseKGroup(nextRoundHead, k);return tail;}public static void main(String[] args) {ListNode a = new ListNode(1);ListNode b = new ListNode(2);ListNode c = new ListNode(3);ListNode d = new ListNode(4);ListNode e = new ListNode(5);ListNode f = new ListNode(6);ListNode g = new ListNode(7);ListNode h = new ListNode(8);a.next = b;b.next = c;c.next = d;d.next = e;e.next = f;f.next = g;g.next = h;h.next = null;int k = 5;//System.out.println("第一种：" + reverseKGroup(a, k));System.out.println("第二种：" + reverseKGroup1(a, k));//System.out.println("第三种：" + reverseKGroup2(a, k));}}