LeetCode - 24. Swap Nodes in Pairs

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

Example:

Given , you should return the list as .1->2->3->42->1->4->3

Note:

• Your algorithm should use only constant extra space.
• You may not modify the values in the list's nodes, only nodes itself may be changed.

交換鏈表中的節點，題目不難，須要細心，有兩種解法。

1.遞歸，邏輯清晰。用臨時節點保存交換中的中間節點，以防鏈表斷裂，節點丟失。

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

2.直接循環作，須要一個假的頭節點來保存交換以後的頭節點，一樣須要臨時節點保存交換中的中間節點。

class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null)
            return head;
        ListNode fakeHead = new ListNode(0), pre = fakeHead, temp = null;
        fakeHead.next = head;
        while (pre.next!=null && pre.next.next!=null) {
            temp = pre.next.next;
            pre.next.next = temp.next;
            temp.next = pre.next;
            pre.next = temp;
            pre = temp.next;
        }
        return fakeHead.next;
    }
}

在LeetCode上循環比遞歸耗時減小1ms