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[LeetCode] Reverse Nodes in k-Group

时间:2014-10-29 23:51:47      阅读:266      评论:0      收藏:0      [点我收藏+]

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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

 

/**
 * 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 <= 0) return head;
        
        ListNode *pcount = head;
        ListNode *pter = head;
        ListNode **prev = &head;
        ListNode *pnext = NULL;
        while (pcount != NULL) {
            int count = 0;
            for (; count < k && pcount != NULL; ++count) { //判断够不够k
                pcount = pcount->next;
            }
            if (count == k) {
                ListNode *pnext = NULL;
                ListNode *pend = pter;
                ListNode *pstart = pter;
                while (pter != pcount) {
                    pnext = pter->next;
                    pter->next = pend;
                    pend = pter;
                    pter = pnext;
                }
                *prev = pend;
                prev = &(pstart->next);
                pstart->next = pter;
            }
        }
        
        return head;
    }
};

 

[LeetCode] Reverse Nodes in k-Group

标签:des   style   blog   io   color   ar   for   sp   div   

原文地址:http://www.cnblogs.com/vincently/p/4060742.html

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