[LeetCode] 025. Reverse Nodes in k-Group (Hard) (C++/Java)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

025. Reverse Nodes in k-Group (Hard)

链接：

题目：https://oj.leetcode.com/problems/reverse-nodes-in-k-group/
代码(github)：https://github.com/illuz/leetcode

题意：

把一个链表每 k 个分为一组，每组内进行翻转。
只能用常数级的空间。

分析：

这题比较考验链表的操作，用递归做会比较方便，先找到下一组的节点，把本组反转后再递归处理后面的节点。

代码：

C++:

class Solution {
public:
	ListNode *reverseKGroup(ListNode *head, int k) {
		if (!head || !(head->next) || k < 2)
			return head;
		
		// count k nodes
		ListNode *nextgp = head;
		for (int i = 0; i < k; i++)
			if (nextgp)
				nextgp = nextgp->next;
			else
				return head;

		// reverse
		ListNode *prev = NULL, *cur = head, *next = NULL;
		while (cur != nextgp) {
			next = cur->next;
			if (prev)
				cur->next = prev;
			else
				cur->next = reverseKGroup(nextgp, k);
			prev = cur;
			cur = next;
		}
		return prev;
	}
};

Java:

public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode cur = head;
        int cnt = 0;
        // get next group
        while (cur != null && cnt != k) {
            cur = cur.next;
            cnt++;
        }
        if (cnt == k) {
            cur = reverseKGroup(cur, k);
            // reverse
            while (0 <= --cnt) {
                ListNode tmp = head.next;
                head.next = cur;
                cur = head;
                head = tmp;
            }
            head = cur;
        }
        return head;
    }
}