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LeetCode: Reverse Nodes in k-Group 解题报告
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
解答：
1. 用递归实现，逐层进行反转。遇到最后一个如果个数不为k，再反转一次即可。
View Code2. 使用一个专用的反转函数来进行反转，从头到尾遍历，遍历到K的时候，使用Pre-Next指针的方式进行反转。这个方法比递归更棒。
1 /* 2 * Solution 2: 3 * 使用区间反转的办法, iteration. 4 * */ 5 public ListNode reverseKGroup(ListNode head, int k) { 6 if (head == null) { 7 return null; 8 } 9 10 ListNode dummy = new ListNode(0);11 dummy.next = head;12 13 ListNode cur = head;14 ListNode pre = dummy;15 16 int cnt = 0;17 18 while (cur != null) {19 cnt++;20 if (cnt == k) {21 cnt = 0;22 pre = reverse(pre, cur.next);23 }24 cur = cur.next;25 }26 27 return dummy.next;28 }29 30 /**31 * Reverse a link list between pre and next exclusively32 * an example:33 * a linked list:34 * 0->1->2->3->4->5->635 * | | 36 * pre next37 * after call pre = reverse(pre, next)38 * 39 * 0->3->2->1->4->5->640 * | |41 * pre next42 * @param pre 43 * @param next44 * @return the reversed list‘s last node, which is the precedence of parameter next45 */46 private static ListNode reverse(ListNode pre, ListNode next){47 ListNode cur = pre.next;48 49 // record the new tail.50 ListNode last = cur;51 while (cur != next) {52 ListNode tmp = cur.next;53 cur.next = pre.next;54 pre.next = cur;55 cur = tmp;56 }57 58 last.next = next;59 return last;60 }主页君的GitHub代码
ref: http://www.cnblogs.com/lichen782/p/leetcode_Reverse_Nodes_in_kGroup.html
LeetCode: Reverse Nodes in k-Group 解题报告
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