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.

first 是第 m -1 个元素。 elementM是第m个元素。 后面将 cur -> second 变成 cur <- second。 最后 cur是第n个元素，second是第n + 1个元素。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ListNode header = new ListNode(-1);
        header.next = head;
        ListNode cur = header;
        ListNode first = null;
        ListNode elementM = null;
        ListNode second = null;
        int i = 0;
        while(i < m){
            first = cur;
            cur = cur.next;
            i ++;
        }
        elementM = cur;
        second = cur.next;
        while(i < n){
            ListNode tmp = second.next;
            second.next = cur;
            cur = second;
            second = tmp;
            i ++;
        }
        first.next = cur;
        if(elementM != null && elementM != second) elementM.next = second;
        return header.next;
    }
}

第三遍：

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode header = new ListNode(-1);
        header.next = head;
        ListNode first = header, second = header;
        for(int i = 1; i < m; i ++){
            first = first.next; 
        }
        ListNode cur = first.next;
        for(int i = 0; i < n; i ++){
            second = second.next;
        }
        ListNode end = second;
        second = second.next;
        end.next = null;
        while(cur != null){
            ListNode tmp = cur.next;
            cur.next = second;
            second = cur;
            cur = tmp;
        }
        first.next = end;
        return header.next;
    }
}