98.Sort List-链表排序（中等题）

链表排序算法

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#LintCode #链表 #归并排序 #快速排序

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本文介绍如何使用归并排序和快速排序两种算法，在O(nlogn)时间复杂度和常数级空间复杂度下对链表进行排序。通过具体示例展示了每种方法的实现过程。

链表排序

题目

在 O(n log n) 时间复杂度和常数级的空间复杂度下给链表排序。
样例

给出 1->3->2->null，给它排序变成 1->2->3->null.
挑战

分别用归并排序和快速排序做一遍。
题解

1.归并排序

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution
{
    public ListNode sortList(ListNode head)
    {
        if (head == null || head.next == null)
        {
            return head;
        }

        ListNode mid = findMiddle(head);

        ListNode right = sortList(mid.next);
        mid.next = null;
        ListNode left = sortList(head);

        return merge(left, right);
    }

    private ListNode findMiddle(ListNode head)
    {
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null)
        {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }

    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;
    }
}

2.快速排序

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution
{
    public ListNode sortList(ListNode head)
    {
        if (head == null || head.next == null)
        {
            return head;
        }

        ListNode mid = findMedian(head); // O(n)

        ListNode leftDummy = new ListNode(0), leftTail = leftDummy;
        ListNode rightDummy = new ListNode(0), rightTail = rightDummy;
        ListNode middleDummy = new ListNode(0), middleTail = middleDummy;
        while (head != null)
        {
            if (head.val < mid.val)
            {
                leftTail.next = head;
                leftTail = head;
            }
            else if (head.val > mid.val)
            {
                rightTail.next = head;
                rightTail = head;
            }
            else
            {
                middleTail.next = head;
                middleTail = head;
            }
            head = head.next;
        }

        leftTail.next = null;
        middleTail.next = null;
        rightTail.next = null;

        ListNode left = sortList(leftDummy.next);
        ListNode right = sortList(rightDummy.next);

        return concat(left, middleDummy.next, right);
    }

    private ListNode findMedian(ListNode head)
    {
        ListNode slow = head, fast = head.next;
        while (fast != null && fast.next != null)
        {
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    private ListNode concat(ListNode left, ListNode middle, ListNode right)
    {
        ListNode dummy = new ListNode(0), tail = dummy;

        tail.next = left;
        tail = getTail(tail);
        tail.next = middle;
        tail = getTail(tail);
        tail.next = right;
        tail = getTail(tail);
        return dummy.next;
    }

    private ListNode getTail(ListNode head)
    {
        if (head == null)
        {
            return null;
        }

        while (head.next != null)
        {
            head = head.next;
        }
        return head;
    }
}