[LeetCode] 24. Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

You may not modify the values in the list's nodes. Only nodes itself 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

两两交换链表中的节点。给一个linked list，请成对调换node。

我的思路是迭代。依然是给一个dummy节点放在head节点之前，然后dummy.next是head节点，nextStart是第三个节点，需要交换的只是l2和l2.next，l1永远是需要swap的两个节点之前的那个节点，他不参与swap。每次交换完了之后只往前挪动一个节点的距离。我画了一个示意图，这样不容易错。

dummy -> 1 -> 2 -> 3 -> 4

l1              l2            ns

时间O(n)

空间O(1)

JavaScript实现

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var swapPairs = function(head) {
    // corner case
    if (head === null || head.next === null) {
        return head;
    }

    // normal case
    let dummy = new ListNode(0);
    dummy.next = head;
    let l1 = dummy;
    let l2 = head;
    while (l2 !== null && l2.next !== null) {
        let nextStart = l2.next.next;
        l1.next = l2.next;
        l2.next.next = l2;
        l2.next = nextStart;
        l1 = l2;
        l2 = l2.next;
    }
    return dummy.next;
};

Java实现

class Solution {
    public ListNode swapPairs(ListNode head) {
        // corner case
        if (head == null || head.next == null) {
            return head;
        }

        // normal case
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode l1 = dummy;
        ListNode l2 = head;
        while (l2 != null && l2.next != null) {
            ListNode nextStart = l2.next.next;
            l1.next = l2.next;
            l2.next.next = l2;
            l2.next = nextStart;
            l1 = l2;
            l2 = l2.next;
        }
        return dummy.next;
    }
}

LeetCode 题目总结