Problem 24: Swap Nodes in Pairs
思路
考察的主要就是两根指针的移动,和上一题很像
/**
* 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) {
ListNode dummy = new ListNode(0);
dummy.next = head;
head = dummy;
while (head.next != null && head.next.next != null) {
ListNode n1 = head.next;
ListNode n2 = head.next.next;
// swap
head.next = n2;
n1.next = n2.next;
n2.next = n1;
head = n1;
}
return dummy.next;
}
}
易错点
dummy node 必须要存在,因为涉及到交换 node 的时候,“头”就乱了。
while 的循环条件,是 head.next.next != null; 因为要涉及两根指针 n1 和 n2
交换完指针记得 update head 指针
// swap
head.next = n2;
n1.next = n2.next;
n2.next = n1;
head = n1;
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