第六周 Reverse Nodes in k-Group

Reverse Nodes in k-Group

LeetCode上的25题：https://leetcode.com/problems/reverse-nodes-in-k-group/submissions/

题目

给出一个链表，和一个整数k，k小于等于链表长度，将链表划分为长度为k的区域，翻转每个区域的元素，若最后一个区域长度不为k，则不对该区域操作。返回翻转后的链表

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

要求：
只允许常数大小的额外空间
不修改node里的value值，只改变node顺序

题解

这里的解法思路比较简单，就是按顺序对每个长度为k的区间进行翻转，直到找到结尾。若链表大小不为k的倍数，则不处理最后一个区间

代码

/**
 * 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 || k < 2) return head;
        ListNode* start = head;
        ListNode* prev_start = new ListNode(0);
        prev_start->next = head;
        ListNode* prev = prev_start;

        while(start != NULL) {
        	ListNode* cur_head = start;
	        ListNode* k_node = start;
            int i;
	        for (i = 0; i < k; i++) {
	        	if (k_node->next == NULL && i != k - 1) break;  //找到尾部且链表长度不为k的倍数
	        	k_node = k_node->next;
	        }
            if (i != k) break;
            
	        start = k_node; //下一个区间的头部

			//翻转过程
	        for (i = 1; i < k; i++) {
	        	ListNode* next_head = cur_head->next;
	        	cur_head->next = k_node;
	        	k_node = cur_head;
	        	cur_head = next_head;
	        }
            
            cur_head->next = k_node;
            ListNode* temp = prev->next;
	        prev->next = cur_head;
	        prev = temp;
            
            ListNode* p = prev_start->next;
            while(p != NULL){
                cout << p->val << " ";
                p = p->next;  
            }
            cout << endl;
        }

        head = prev_start->next;
        delete prev_start;
        return head;
    }
};