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.
题意 : 交换相邻两个节点
此题不难,但是出错的地方 就在于 不要把引用搞混了,p = q 把q赋给p,此时p也指向同一块内存
如果此时p 对 这块内存的修改,q也能够看见 ,但是如果p = t ,那么等价的就是p现在指向了另一块内存
此时你觉得p的影响会影响到q吗,明显不会,这就是为什么要建立父指针的原因所在!
public class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) return head;
ListNode parent = head;
ListNode son = head;
int count = 0;
while (son != null && son.next != null) {
ListNode tmp = son.next;
son.next = tmp.next;
tmp.next = son;
if (count == 0) {
head = tmp;
count = 1;
} else {
parent.next = tmp;
}
parent = tmp.next;
son = parent.next;
}
return head;
}
}