LeetCode 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

思路：递归的处理，然后每层做group大小的翻转就是了：一个pre为前一位指针，然后还一个cur指针，再一个next指针就行了，每次就只翻转一个节点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if (head == NULL || head->next == NULL || k <= 1) return head;

        int n = k, len = 0;
        ListNode *cur = head;
        while (cur != NULL) {
            ++len;
            cur = cur->next;
        }
        if (n > len) return head;

        cur = head;
        ListNode *pre = NULL;
        while (cur != NULL && n > 0) {
            ListNode *ne = cur->next;
            cur->next = pre;
            pre = cur;
            cur = ne;
            n--;
        }

        if (len - k >= k) 
            head->next = reverseKGroup(cur, k);
        else head->next = cur;

        return pre;
    }
};

LeetCode Reverse Nodes in k-Group

原文：http://blog.csdn.net/u011345136/article/details/43854825