Leetcode 25 Reverse Nodes in k-Group 插入到每组头部，链表容易插

最新推荐文章于 2022-04-08 22:28:12 发布

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本文介绍了一种算法，该算法将链表中的节点每K个一组进行反转，并返回修改后的链表。讨论了当节点数不是K的倍数时的处理方式，并提供了两种实现方法，包括简化版代码示例。

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

---------------------------------------------------------------------

从discussion学的更加简化的版本：

-1 -> 1 -> 2 -> 3 -> 4 -> 5
 |    |    |    | 
pre  cur  nex  tmp

-1 -> 2 -> 1 -> 3 -> 4 -> 5
 |         |    |    | 
pre       cur  nex  tmp

-1 -> 3 -> 2 -> 1 -> 4 -> 5
 |              |    |    | 
pre            cur  nex  tmp

class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
        if(head==NULL||k==1) return head;
        int num=0;
        ListNode *preheader = new ListNode(-1);
        preheader->next = head;
        ListNode *cur = preheader, *nex, *pre = preheader;
        while(cur = cur->next) 
            num++;
        while(num>=k) {
            cur = pre->next;
            nex = cur->next;
            for(int i=1;i<k;++i) {
                cur->next=nex->next;
                nex->next=pre->next;
                pre->next=nex;
                nex=cur->next;
            }
            pre = cur;
            num-=k;
        }
        return preheader->next;
    }
};

My codes before several years:

1. Counting the length is much easier than reversing the last less than k nodes.

2. Don't forget: lastGroupTail->next = p;

/**
 * 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) {
    // IMPORTANT: Please reset any member data you declared, as
    // the same Solution instance will be reused for each test case.
    if (k <= 1 || head == NULL)
      return head;
    ListNode* dummy = new ListNode(0);
    dummy->next = head;
    ListNode *p = head, *last = NULL, *next = NULL, *lastGroupTail = dummy, *curGroupTail = NULL, *res = NULL;
    int count = 1, len = 0;
    for(p = head; p != NULL; p = p->next, ++len);
    p = head;
    int countmax = len / k * k;
    while (p != NULL && count <= countmax) {
      next = p->next;
      if (count % k == 1)
        curGroupTail = p;
      else if (count % k == 0) {
        lastGroupTail->next = p;
        lastGroupTail = curGroupTail;
      }
      p->next = last;
      ++count;
      last = p;
      p = next;
    }
    lastGroupTail->next = p;
    res = dummy->next;
    return res;
  }
};