22 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.

 

把一个链表中的每一对节点对换（不能只换值）

/**
 * 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 h = new ListNode(0);
    h.next = head;
    ListNode p = h;
 
    while(p.next != null && p.next.next != null){
        //use t1 to track first node
        ListNode t1 = p;
        p = p.next;
        t1.next = p.next;
 
        //use t2 to track next node of the pair
        ListNode t2 = p.next.next;
        p.next.next = p;
        p.next = t2;
    }
 
    return h.next;
}
}

 

 

解题思路：这题主要涉及到链表的操作，没什么特别的技巧，注意不要出错就好。最好加一个头结点，操作起来会很方便。

 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param a ListNode
    # @return a ListNode
    def swapPairs(self, head):
        if head == None or head.next == None:
            return head
        dummy = ListNode(0); dummy.next = head
        p = dummy
        while p.next and p.next.next:
            tmp = p.next.next
            p.next.next = tmp.next
            tmp.next = p.next
            p.next = tmp
            p = p.next.next
        return dummy.next