LeetCode刷题-#25 Reverse Nodes in k-Group

描述

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

分析

无

代码（C++）

// 方法一
// LeetCode, Reverse Nodes in k-Group
// 递归版，时间复杂度 O(n)，空间复杂度 O(1)
class Solution{
public:
    ListNode *reverseKGroup(ListNode *head, int k){
	if(head == nullptr || head->next == nullptr || k < 2)
	    return head;
	
	ListNode *next_group = head;
	for(int i = 0; i < k; ++i){
	    if(next_group)
		next_group = next_group->next;
	    else
		return head;
	}
	
	// next_group is the head of next group
	// new_next_group is the new head of next group after reversion
	ListNode *new_next_group = reverseKGroup(next_group, k);
	ListNode *prev = NULL, *cur = head;
	while(cur != next_group){
	    ListNode *next = cur->next;
	    cur->next = prev ? prev : new_next_group;
	    prev = cur;
	    cur = next;
	}
	return prev;	// prev will be the new head of this group
    }
};

相关题目

• LeetCode刷题-#24 Swap Nodes in Pairs

标签：25,head,Group,Reverse,next,return,group,prev,cur
来源： https://www.cnblogs.com/sdyan/p/16372201.html