LinkedList - 148. Sort List

148. Sort List

Sort a linked list in O(n log n) time using constant space complexity.

Example 1:

Input: 4->2->1->3
Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0
Output: -1->0->3->4->5

思路：

题目意思很明确，给一个链表排序，排序的方式有很多，这里写一下归并排序。

代码：

java：

/**

 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head;
        
        // 1. find mid of list
        ListNode fast = head, slow = head, partHead = slow;
        while (fast != null && fast.next != null) {
            partHead = slow;
            fast = fast.next.next;
            slow = slow.next;
        }
        
        // 2. cut list as binary
        partHead.next = null;
        
        // 3. recursion cut
        ListNode left = sortList(head);
        ListNode right = sortList(slow);
        
        // 4. merge list
        return merge(left, right);
    }
    
    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode res = new ListNode(0), p = res;
        
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        
        if ( l1 != null) {
            p.next = l1;
        }
         
        if (l2 != null) {
            p.next = l2;
        }
        
        return res.next;
    }
}