[数据结构与算法]【咸鱼刷LeetCode-24】两两交换链表中的节点(中等)

解法一：交换指针

public ListNode swapPairs(ListNode head) {
    	if (head == null || head.next == null) {
			return head;
		}
    	ListNode newHead = head.next;
    	ListNode cur = head;
    	ListNode next = cur.next;
    	ListNode temp = null;
    	while (next != null) {
    		temp = next.next;
    		cur.next = next.next;
    		next.next = cur;
    		
    		if (temp == null) {
    			break;
			}
    		
    		if (temp.next != null) {
    			cur.next = temp.next;
			}
    		cur = temp;
    		next = cur.next;
		}
    	return newHead;
    }

解法二：交换value

因为给出的ListNode有value值，其实只交换Node的value值更简单，不必处理指针的问题，两者执行结果也差不多：

public ListNode swapPairs(ListNode head) {
    	if (head == null || head.next == null) {
			return head;
		}
    	ListNode cur = head;
    	ListNode next = cur.next;
    	while (next != null) {
			cur.val = cur.val + next.val;
			next.val = cur.val - next.val;
			cur.val = cur.val - next.val;
			
			cur = next.next;
			next = cur;
			if (null != cur) {
				next = cur.next;
			} 
		}
    	return head;
    }

?