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#yyds干货盘点# leetcode算法题： K 个一组翻转链表
source link: https://blog.51cto.com/u_13321676/5372357
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#yyds干货盘点# leetcode算法题： K 个一组翻转链表
原创给你链表的头节点 head ，每 k 个节点一组进行翻转，请你返回修改后的链表。
k 是一个正整数，它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍，那么请将最后剩余的节点保持原有顺序。
你不能只是单纯的改变节点内部的值，而是需要实际进行节点交换。
输入：head = [1,2,3,4,5], k = 2
输出：[2,1,4,3,5]
输入：head = [1,2,3,4,5], k = 3
输出：[3,2,1,4,5]
代码实现：
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if(head == null || head.next == null) return head;
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode cur = head;
int count = 0;
while(cur != null)
{
cur = cur.next;
count++;
}
int timesToFlip = count / k;
cur= head;
ListNode before = dummyHead;
ListNode after = null;
for(int i = 0; i < timesToFlip; i++)
{
int numNodes = k;
ListNode temp = null, pre = null;
//Reverse partial Linked-list
while(numNodes != 0)
{
numNodes--;
temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
after = cur;
}
before.next = pre;
while(pre.next != null)
pre = pre.next;
pre.next = after;
before = pre;
cur = before.next;
}
return dummyHead.next;
}
}
public ListNode reverseKGroup(ListNode head, int k) {
if(head == null || head.next == null) return head;
ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode cur = head;
int count = 0;
while(cur != null)
{
cur = cur.next;
count++;
}
int timesToFlip = count / k;
cur= head;
ListNode before = dummyHead;
ListNode after = null;
for(int i = 0; i < timesToFlip; i++)
{
int numNodes = k;
ListNode temp = null, pre = null;
//Reverse partial Linked-list
while(numNodes != 0)
{
numNodes--;
temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
after = cur;
}
before.next = pre;
while(pre.next != null)
pre = pre.next;
pre.next = after;
before = pre;
cur = before.next;
}
return dummyHead.next;
}
}
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