題目連結

提交程式碼：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
	public ListNode reverseKGroup(ListNode head, int k) {
		if (head == null || k == 1)
			return head;
		ListNode newhead =
 
 new ListNode(-1);
		newhead.next = null;
		int i, cnt = 0;

		ListNode pmark = head;
		ListNode grouphead = newhead;
		ListNode p = head;
		ListNode pnext = p.next;
		ListNode nextgrouphead = p;
		while (pmark != null) {
			for (i = 0; i < k && pmark != null; i++) {
				// pmark會移動到下一個翻轉小組的第一個位置
 

				pmark = pmark.next;
				cnt++;
			}

			// 退出迴圈之後如果cnt==k則翻轉小組中的結點個數有k個需要反轉
			if (cnt == k) {
				for (i = 0; i < k; i++) {
				//頭插法
					p.next = grouphead.next;
					grouphead.next = p;
					p = pnext;
					if (p != null)
						pnext = p.next;
				}
				grouphead = nextgrouphead;
				nextgrouphead =
 
 p;
				cnt = 0;
			} else {
				grouphead.next = p;
				return newhead.next;
			}

		}
		return newhead.next;

	}
}

執行結果：