92. Reverse Linked List II - jiejackyzhang/leetcode-note GitHub Wiki

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.

解题思路为：

1. 设置dummy node，防止head丢失；
2. 找到需要reverse之前的node，设为prev；
3. 取得需要reverse的第一个node，和它的next的node，通过for-loop依次reverse；
4. 最后返回dummy.next。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if(head == null) return null;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        for(int i = 0; i < m-1; i++) {
            pre = pre.next;
        }
        ListNode cur = pre.next;
        for(int i = 0; i < n-m; i++) {
            ListNode temp = cur.next;
            cur.next = temp.next;
            temp.next = pre.next;
            pre.next = temp;
        }
        return dummy.next;
    }
}