题解 | 链表内指定区间反转

解法一：

public ListNode reverseBetween (ListNode head, int m, int n) {
        // write code here
        if (head == null || m == n) return head; // 如果链表为空或反转区间只有一个节点

        ListNode pre = null, cur = head;

        // 1. 找到第 m-1 个节点 (pre) 和第 m 个节点 (cur)
        for (int i = 1; i < m; i++) {
            pre = cur;
            cur = cur.next;
        }

        // 2. 保存第 m 个节点前后的节点
        ListNode lastEnd = pre; // 记录第 m-1 个节点，反转后连接这里
        ListNode firstNode = cur; // 第 m 个节点，将成为反转后的最后一个节点

        // 3. 反转区间 [m, n] 内的节点
        ListNode next = null;
        for (int i = m; i <= n; i++) {
            next = cur.next;  
            cur.next = pre;  
            pre = cur;     
            cur = next;      
        }

        // 4. 两种情况，一种是m不为1，代表前面还有节点；另外一种代表m为1，就是第一个节点
        if (lastEnd != null) {
            lastEnd.next = pre; // 连接第 m-1 个节点与反转后的部分
        } else {
            head = pre; // 如果 m == 1，那么反转后的头节点为 pre
        }

        // 5. 将反转区间最后一个节点（即为反转区间的第m个节点）的 next 指向第 n+1 个节点
        firstNode.next = cur;

        return head;
}

解法二（思路来源于左神所讲的划分链表）：

public ListNode reverseBetween (ListNode head, int m, int n) {
        // write code here
        if (head == null || head.next == null || m == n) return head;

        ListNode p = head;
        ListNode lastEnd = null, curStart = null, curEnd = null, afterStart = null;
        int l = m, r = n;
        while (r > 0) {
            l--;
            r--;
            if (l > 0) {
                lastEnd = p;
            }
            if (l == 0) {
                curStart = p;
            }
            if (r == 0) {
                curEnd = p;
            }
            p = p.next;
        }

        ListNode cur = curStart;
        afterStart = curEnd.next;
        curEnd.next = null;

        ListNode pre = null, next = null;
        while (cur != null) {
            next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }

        if (lastEnd != null) {
            lastEnd.next = pre;
        }
        if (afterStart != null) {
            curStart.next = afterStart;
        }

        head = lastEnd == null? pre : head;
        return head;
    }