这道题目是链表倒转的重要总结

https://leetcode.com/problems/reverse-nodes-in-k-group/?tab=Description

 

解答：

https://discuss.leetcode.com/topic/7126/short-but-recursive-java-code-with-comments

 

关于链表倒转，开始我都是想加一个dummy node，但是后来发现，也不是完全必要。

 

看上面的解法，就处理的非常简洁，用三个指针，

a->b->c

 

head = a;

cur = null;

tmp = head->next;

head->next = cur;

cur = head;

head = tmp;

这时候，就变成了 a->null, b->c，并且cur指向a，head指向b，

然后一直走到 tmp == null的时候，就不需要把head变成tmp了，直接返回head就可以了。

或者，最后head是null的时候，把cur返回就可以了，因为cur指向的是上一次的head，这也是原解法中的方式。

 

public ListNode reverseKGroup(ListNode head, int k) {
    ListNode curr = head;
    int count = 0;
    while (curr != null && count != k) { // find the k+1 node
        curr = curr.next;
        count++;
    }
    if (count == k) { // if k+1 node is found
        curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
        // head - head-pointer to direct part, 
        // curr - head-pointer to reversed part;
        while (count-- > 0) { // reverse current k-group: 
            ListNode tmp = head.next; // tmp - next head in direct part
            head.next = curr; // preappending "direct" head to the reversed list 
            curr = head; // move head of reversed part to a new node
            head = tmp; // move "direct" head to the next node in direct part
        }
        head = curr;
    }
    return head;
}