Swap Nodes in Pairs

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.

 这道题如果直接交换值是很easy的

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(null == head || null == head.next)
            return head;
        ListNode front = head;
        ListNode behind = front.next;
        
        do{
            int temp = front.val;
            front.val = behind.val;
            behind.val = temp;
            if(null == behind.next || null == behind.next.next)
                break;
            front = behind.next;
            behind = front.next;
        }while(true);
        
        return head;
    }
}

 按照题目要求不交换值，交换节点的引用

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if(null == head || null == head.next)
            return head;
        ListNode behind = head;
        ListNode front = behind.next;
        ListNode temp;
        ListNode pre;
        //交换开头两个节点
        
        behind.next = front.next;
        front.next = behind;
        
        head = front;
      //交换一下front和behind
        temp = behind;
        behind = front;
        front = temp;
        
        pre = front;//倒数第二个指针
        
        while(null != front.next && null != front.next.next){
            behind = front.next;
            front = behind.next;
            
            behind.next = front.next;
            front.next = behind;
            
            temp = behind;
            behind = front;
            front = temp;
            
            pre.next = behind;
            pre = front;
        }
        
        return head;
    }
}