[LeetCode]92. Reverse Linked List II反转部分链表

/*
    重点还是反转链表
    思路就是中间的反转,然后两头接上
     */
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head==null||m>=n) return head;
        int count = 1;
        ListNode sta = head;
        //mid就是第一个接点的前节点
        ListNode mid = null;
        while (count<m)
        {
            mid = head;
            head = head.next;
            count++;
        }
        //中间的部分反转,此时的head是反转的开头
        ListNode pre = null;
        while (count<=n)
        {
            ListNode next = head.next;
            head.next = pre;
            pre = head;
            head = next;
            count++;
        }
        //将后边的接上
        ListNode end = pre;
        while (end.next!=null) end = end.next;
        end.next = head;
        //如果是从第一个开始,要单独考虑
        if (mid!=null) mid.next = pre;
        else return pre;
        return sta;
    }
原文地址:https://www.cnblogs.com/stAr-1/p/8423551.html