92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

Solution 1：Iterative 找到位置后 依次reverse

思路可参考206题1a：http://www.jianshu.com/p/56353df45769
找到reverse前的位置，其余和206题1a相同，再link上即可

屏幕快照 2017-09-12 下午12.44.10.png

Time Complexity: O(N) Space Complexity: O(1)

Solution 2：Iterative 找到位置后 (前部)插入法reverse

思路可参考206题1b：http://www.jianshu.com/p/56353df45769
思路：前部插入法reverse: 1 -> 2 -> 3 -> 4
2 -> 1 -> 3 -> 4 再 3 -> 2 -> 1 -> 4 再 4 -> 3 -> 2 -> 1
找到reverse前的位置，其余和206题1b相同，再link上即可

屏幕快照 2017-09-12 下午12.29.26.png

Time Complexity: O(N) Space Complexity: O(1)

Solution1 Code：

class Solution1 {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null) return null;
        if(m == n) return head;  // just in case
        
        ListNode dummy = new ListNode(0); 
        dummy.next = head;
        
        // find before position, which at the node before reversing
        ListNode before = dummy;
        for(int i = 0; i < m-1; i++) before = before.next; 

        // init
        ListNode r_head = before.next; // a pointer to the beginning of a sub-list that will be reversed
        ListNode cur = r_head;
        ListNode prev = null; //(prev to cur every time)

        // start reversing
        for(int i = 0; i < n - m + 1; i++)
        {
            ListNode next = cur.next;
            cur.next = prev;
            prev = cur;
            cur = next;
        }
        
        // relink 
        before.next = prev;
        r_head.next = cur;

        return dummy.next;
    }
}

Solution2 Code：

class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null) return null;
        if(m == n) return head;  // just in case
        
        ListNode dummy = new ListNode(0); 
        dummy.next = head;
        
        // find before position, which at the node before reversing
        ListNode before = dummy;
        for(int i = 0; i < m-1; i++) before = before.next; 

        // init
        ListNode r_head = before.next; // a pointer to the beginning of a sub-list that will be reversed
        ListNode cur = r_head;
        ListNode post = null; //(right after cur every time)

        // start reversing
        for(int i = 0; i < n - m; i++)
        {
            post = cur.next;
            cur.next = cur.next.next;
            post.next = r_head;
            r_head = post;
        }
        
        // relink r_head
        before.next = r_head;

        return dummy.next;
    }
}