Description
Give a linked list, eachKNodes in a group, and return the flipped linked list.
KIs a positive integer whose value is smaller than or equal to the length of the linked list. If the total number of nodes is notKTo keep the remaining nodes in the original order.
Example:
Given the linked list:1->2->3->4->5
WhenK= 2, should return:2->1->4->3->5
WhenKWhen it is 3, it should return:3->2->1->4->5
Note:
- Your algorithm can only use extra space of constants.
- Instead of simply changing the internal value of a node, You need to actually exchange nodes.
Solutions
First, traverse the entire linked list to get the total length, and then reverse each K nodes in turn. Each time you save the first and last nodes after the current node is reversed, you can splice the linked list of up and down K nodes; record the first node of the new linked list after the first reversal. If the last node is less than K, splice the End Node of the last reverse linked list to the subsequent node.
Code
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode* reverseKGroup(ListNode* head, int k) {12 ListNode *left = head, *right = head, *newHead = NULL;13 int len = 0;14 while(right){15 len++;16 right = right->next;17 }18 if(len < k) return head;19 right = head;20 while(len >= k){21 ListNode *node = right;22 ListNode *h = right;23 right = right->next;24 for(int i = 0; i < k-1; i++){25 ListNode *next = right->next;26 right->next = node;27 node = right;28 right = next;29 }30 if(newHead == NULL) newHead = node;31 else{32 left->next = node;33 left = h;34 }35 len -= k;36 }37 left->next = right;38 return newHead;39 }40 };
Leetcode 25. k a group of flip linked lists (reverse nodes in K-group)