Given a linked list, reverse the nodes of a linked listkat a time and return its modified list.
If the number of nodes is not a multiple ofkthen 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
Fork= 2, you should return:2->1->4->3->5
Fork= 3, you should return:3->2->1->4->5
题意：给你一个链表，让你按group的大小翻转链表。
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;}};
未曾失败的人恐怕也未曾成功过。
