[leetcode] Sort List

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

https://oj.leetcode.com/problems/sort-list/

思路：归并排序，递归执行。

public class Solution {
    public ListNode sortList(ListNode head) {
        if(head==null||head.next==null)
            return head;
        ListNode fast =head;
        ListNode slow =head;
        while(fast!=null&&fast.next!=null&&fast.next.next!=null){
            fast=fast.next.next;
            slow=slow.next;
        }
        ListNode one =head;
        ListNode two =slow.next;
        slow.next=null;
        one = sortList(one);
        two = sortList(two);
        
        return merge(one,two);
        
    }
    
    private ListNode merge(ListNode l1, ListNode l2){
        if(l1==null)
            return l2;
        if(l2==null)
            return l1;
        ListNode dummyHead = new ListNode(-1);
        ListNode p =dummyHead;
        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 dummyHead.next;
        
    }
    
}

快速排序

public class Solution {
    public ListNode sortList(ListNode head) {
        ListNode p = head;
        while (p != null)
            p = p.next;
        quickSort(head, null);
        return head;
    }

    private void quickSort(ListNode from, ListNode to) {
        if (from != to) {
            ListNode p = partition(from, to);
            quickSort(from, p);
            quickSort(p.next, to);
        }

    }

    private ListNode partition(ListNode from, ListNode to) {
        int key = from.val;
        ListNode p = from;
        ListNode q = p.next;

        while (q != to) {
            if (q.val <= key) {
                p = p.next;
                int tmp = p.val;
                p.val = q.val;
                q.val = tmp;
            }
            q = q.next;
        }
        int tmp = p.val;
        p.val = from.val;
        from.val = tmp;

        return p;
    }

    public static void main(String[] args) {
        ListNode head = ListUtils.makeList(5, 4, 1, 6, 2, 1);
        new Solution().sortList(head);
        ListUtils.printList(head);
        Object obj;
    }
}