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  • [LeetCode] 24. Swap Nodes in Pairs ☆

    Given a linked list, swap every two adjacent nodes and return its head.

    For example,

    Given 1->2->3->4, you should return the list as 2->1->4->3.

    Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

    解法:

      循环的方式:

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ListNode swapPairs(ListNode head) {
            ListNode dummy = new ListNode(0);
            dummy.next = head;
            
            head = dummy;
            while (head.next != null && head.next.next != null) {
                ListNode left = head.next;
                ListNode right = left.next;
                
                // head -> left -> right -> ....
                // to: head -> right -> left -> ....
                left.next = right.next;
                right.next = left;
                head.next = right;
                head = left;
            }
            
            return dummy.next;
        }
    }

      递归的方式:

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */
    public class Solution {
        public ListNode swapPairs(ListNode head) {
            if (head == null || head.next == null) {
                return head;
            }
            
            ListNode temp = head.next;
            head.next = swapPairs(temp.next);
            temp.next = head;
            return temp;
        }
    }
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  • 原文地址:https://www.cnblogs.com/strugglion/p/6419363.html
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