找LeetCode上简单的题目来撑过30天啦(DAY24)

好像有台风要来，天气变得有点冷

题号:24 标题：Swap Nodes in Pairs 难度：Medium

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.)

Example 1:

Input: head = [1,2,3,4]
Output: [2,1,4,3]

Example 2:
Input: head = []
Output: []

Example 3:
Input: head = [1]
Output: [1]

Constraints:
• The number of nodes in the list is in the range [0, 100].
• 0 <= Node.val <= 100

我的程序码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode swapPairs(ListNode head) {
        if(head==null || head.next == null){
            return head;
        }else{
            ListNode f = new ListNode();
            ListNode s = new ListNode();
            ListNode c = new ListNode();
            ListNode temp = new ListNode();
            f = head;
            s = head.next;
            if(head.next==null){
                head =head.next;
                head.next = f;
                return head;
            }
            head = head.next;
            c= head.next;
            //System.out.print("i: "+ c.val+" ");
            int i =1;
            while(s!=null){
                System.out.print(i+": "+ s.val+" ");
                s.next = f;
                temp.next = s;
                if(c!=null){
                    f.next = c;
                    c = c.next;
                    if(c != null){
                        c = c.next;    
                    }
                    
                }else{
                    f.next = null;
                }
                temp = f;
                f = f.next;
                s = s.next;
                
                s = s.next;
                if(s == null){
                    break;
                }
                
                s = s.next;
                
//                 System.out.print(i+": "+ cur.val+" ");
                
//                 System.out.print(i+": "+(cur.next).val +" ");
                
//                 System.out.println(i+": "+(p).val);
            }
        }
        
        return head;
    }
}

逻辑
把前一个指向现在的下一个，现在的指向前一个。

DAY24心得
我觉得我跟LINKLIST真的有比较熟了，开心