(java)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.


解题思路:

就是不断地交换相邻两个节点的next,不断循环下去直到结束,记得就是在交换当前两个节点的next(例如k跟k+1)时,记得更新k-1的next。


/**
 * 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;
        //当前节点k
        ListNode p = head;
        //下一节点k+1
        ListNode q = head.next;
        //记录上一节点k-1
        ListNode r = null;
        head = q;
        while(p != null && q != null)
        {
            //交换next
        	p.next = q.next;
        	q.next = p;
        	//更新上一节点的next
        	if(r != null)
        		r.next = q;
        	//更新节点
        	r = p;
        	p = p.next;
        	if(p != null)
        		q = p.next;
        }
        return head;
    }
}


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