Leetcode 24. Swap Nodes in Pairs

思路：添加头节点，依次反转相邻元素，保持反转后的最后一个指针pre，当前被反转的第一个元素的指针cur，当前被反转的第二个元素的指针next（如果存在的话）。反转的思路和92. Reverse Linked List II差不多，只不过pre要移动。

迭代做法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null) return head;
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy, cur = head;
        while(cur != null && cur.next != null) {//保证有2个可以被反转的元素
            ListNode next = cur.next;
            cur.next = next.next;
            next.next = cur;
            pre.next = next;
            pre = cur;
            cur = cur.next;
        }
        return dummy.next;
    }
}

 

递归做法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head == null || head.next == null) return head;//保证有2个可以被反转的元素
        ListNode cur = head, next = cur.next;
        cur.next = swapPairs(next.next);
        next.next = cur;
        return next;
    }
}

 

Next challenges: Reverse Nodes in k-Group