leetcode刷題-148. Sort List

Sort a linked list in O(n log n) time using constant space complexity.
Input: 4->2->1->3
Output: 1->2->3->4
Input: -1->5->3->4->0
Output: -1->0->3->4->5

這道題是對列表進行排序，要求時間複雜度是nlogn，我用的是歸併排序，和數組歸併排序不同的是找到中點，在列表中通過快慢指針找中點，當快指針到達終點時慢指針到達中點，然後就是一般的歸併排序思想

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        //遞歸出口
        if (head == null || head.next == null) {
            return head;
        }
        //找到中點
        ListNode middle = findMiddle(head);
        //排序右面
        ListNode right = sortList(middle.next);
        //排序左面
        middle.next = null;
        ListNode left = sortList(head);
        //左右歸併
        return merge(left, right);
    }
    //快慢指針找列表中點
    private ListNode findMiddle(ListNode node) {
        ListNode slow = node, fast = node.next;
        while(fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
    //列表merge操作
    private ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        
        while(head1 != null && head2 != null) {
            if (head1.val < head2.val) {
                tail.next = head1;
                head1 = head1.next;
            } else {
                tail.next = head2;
                head2 = head2.next;
            }
            
            tail = tail.next;
        }
        if (head1 != null) {
            tail.next = head1;
        } else {
            tail.next = head2;
        }
        
        return dummy.next;
    }
}