链表Reverse List

（一）reverse linked list

https://leetcode.com/problems/reverse-linked-list/description/

题目：逆转链表；

解答：使用两个指针:prev和curt, 从头到尾依次变换指针方向；

代码：

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        return prev;
    }
}

（二）reverse linked list ii

https://leetcode.com/problems/reverse-linked-list-ii/description/；

题目：将链表从m到n的位置逆转；

解答：需要讲m前一个节点和n的后一个节点保存，以便逆转完成后重新连接链表；

代码：

class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null || m >= n) {
            return head;
        }  
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        for (int i = 1; i < m; i++) {
            head = head.next;
        }
        ListNode prev = head;
        ListNode cutLeft = prev.next;
        head = head.next;
        ListNode post = head.next;
        
        for (int i = m; i < n; i++) {
            if (head != null && post != null) {
                ListNode temp = post;
                post = temp.next;
                temp.next = head;
                head = temp;
            }
        }
        prev.next = head;
        cutLeft.next = post;
        return dummy.next;
    }
}

（三）reverse nodes in k-group

https://leetcode.com/problems/reverse-nodes-in-k-group/description/

题目：将链表每k组元素进行逆转；

          例如：

         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

解答：写出单个逆转的子函数，形参为当前需逆转链表的前一个节点和k，返回逆转完成后的链表最后一个节点（即为下一个需逆转链表的前一个节点）；

           原链表：head -> n1 -> n2 -> ... -> nk -> nk+1;

           变换后：head -> nk ->nk-1 -> ... -> n1 ->nk+1;

           返回：n1;

代码：

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        while (head != null) {
            head = reverseK(head, k);
        }
        return dummy.next;
    }
    
    private ListNode reverseK(ListNode head, int k) {
        ListNode reversePrev = head;
        ListNode prev = null;
        head = head.next;
        ListNode reverseHead = head;
        ListNode reversePost = head;
        for (int i = 0; i < k; i++) {
            if (reversePost == null) {
                return null;
            }
            reversePost = reversePost.next;
        }
        while (head != reversePost) {
            ListNode temp = head.next;
            head.next = prev;
            prev = head;
            head = temp;
        }
        reversePrev.next = prev;
        reverseHead.next = head;
        return reverseHead;
    }
}