[LeetCode24]Swap Nodes in Pairs-优快云博客

本文链接：https://blog.youkuaiyun.com/sbitswc/article/details/30124845

本文介绍了一种链表中相邻节点的交换算法，通过迭代或递归的方式实现，不使用额外空间，仅改变节点指针指向来完成节点位置的交换。

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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.

Analysis:

Could use recursion or iterative

Java

public ListNode swapPairs(ListNode head) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        ListNode helper = new ListNode(0);
        helper.next = head;
        ListNode n1 = helper;
        ListNode n2 = head;
        while(n2!=null && n2.next!=null){
        	ListNode temp = n2.next.next;
        	n2.next.next = n1.next;
        	n1.next = n2.next;
        	n2.next = temp;
        	n1 = n2;
        	n2 = n1.next;
        }
        return helper.next;
    }

another

public ListNode swapPairs(ListNode head) {
        if(head==null||head.next==null) return head;
        ListNode newHead = new ListNode(-1);
        newHead.next = head;
        ListNode p1 = newHead;
        int len = 0;
        ListNode l1 = head;
        while(l1!=null){
        	l1 = l1.next;
        	len++;
        }
        int dis = len/2;
        l1 = head;
        while(dis>0){
            ListNode l2 = l1.next;
            l1.next = l2.next;
            l2.next = l1;
            p1.next = l2;
            p1 = l1;
            l1 = l1.next;
        	dis--;
        }
        return newHead.next;
    }

c++

recursion

ListNode *swapPairs(ListNode *head) {
       if(head == NULL || head->next == NULL)
        return head;
    ListNode *newPair = head->next->next;
    ListNode *newHead = head->next;
    head->next->next = head;
    head->next = swapPairs(newPair);
    return newHead; 
    }