24. Swap Nodes in Pairs

2016-12-20  本文已影响2人  hyhchaos

Java

/**
 * 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) return null;
        if(head.next==null) return head;
        ListNode tmp=head.next;
        ListNode nt=head;
        head=head.next;
        while(nt.next!=null&&tmp.next!=null)
        {
            ListNode v=tmp.next.next;
            ListNode w=tmp.next;
            tmp.next=nt;
            if(v!=null)
            {
            nt.next=v;
            tmp=v;
            nt=w;
            }
            else
            {
            nt.next=w;
            nt=w;
            }
        }
        if(nt.next==null)
        return head;
        tmp.next=nt;
        nt.next=null;
        return head;
    }
}

优解,Java,方法差不多,但在head节点之前添加一个节点的做法很nice

public ListNode swapPairs(ListNode head) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode current = dummy;
    while (current.next != null && current.next.next != null) {
        ListNode first = current.next;
        ListNode second = current.next.next;
        first.next = second.next;
        current.next = second;
        current.next.next = first;
        current = current.next.next;
    }
    return dummy.next;
}
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