148. 排序链表

给你链表的头结点 head ，请将其按 升序 排列并返回 排序后的链表 。

进阶：

你可以在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序吗？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sort-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

递归

class Solution {

    private ListNode findMid(ListNode head) {
        if (head == null || head.next == null || head.next.next == null) {
            return head;
        }

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

        return slow;
    }

    private ListNode merge(ListNode p1, ListNode p2) {
        ListNode newHead = new ListNode(), tail = newHead;
        while (p1 != null && p2 != null) {
            if (p1.val <= p2.val) {
                tail.next = p1;
                p1 = p1.next;
            } else {
                tail.next = p2;
                p2 = p2.next;
            }
            tail = tail.next;
        }

        if (p1 != null) {
            tail.next = p1;
        }

        if (p2 != null) {
            tail.next = p2;
        }

        return newHead.next;
    }

    private ListNode mergeSort(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode mid = findMid(head);
        ListNode next = mid.next;
        mid.next = null;
        ListNode p1 = mergeSort(head), p2 = mergeSort(next);
        return merge(p1, p2);
    }

    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        return mergeSort(head);
    }

    public static void main(String[] args) {
        ListNode n1 = new ListNode(4);
        ListNode n2 = new ListNode(2);
        ListNode n3 = new ListNode(1);
        ListNode n4 = new ListNode(3);

        n1.next = n2;
        n2.next = n3;
        n3.next = n4;

        ListNode node = new Solution().sortList(n1);
        while (node != null) {
            System.out.println(node.val);
            node = node.next;
        }
    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode() {
    }

    ListNode(int val) {
        this.val = val;
    }

    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}

迭代

class Solution {

    private ListNode findMid(ListNode head) {
        if (head == null || head.next == null || head.next.next == null) {
            return head;
        }

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

        return slow;
    }

    private ListNode merge(ListNode p1, ListNode p2) {
        ListNode newHead = new ListNode(), tail = newHead;
        while (p1 != null && p2 != null) {
            if (p1.val <= p2.val) {
                tail.next = p1;
                p1 = p1.next;
            } else {
                tail.next = p2;
                p2 = p2.next;
            }
            tail = tail.next;
        }

        if (p1 != null) {
            tail.next = p1;
        }

        if (p2 != null) {
            tail.next = p2;
        }

        return newHead.next;
    }

    private ListNode mergeSort(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        int length = 0;
        ListNode cur = head;

        while (cur != null) {
            cur = cur.next;
            length++;
        }

        ListNode newHead = new ListNode(), tail;
        newHead.next = head;

        for (int step = 1; step < length; step <<= 1) {
            tail = newHead;
            cur = newHead.next;
            while (cur != null) {
                int count = 0;
                ListNode next = null;
                ListNode head1 = cur;

                /**
                 * 切分step个节点
                 */
                while (++count < step && cur != null) {
                    cur = cur.next;
                }
                next = cur == null ? null : cur.next;
                if (cur != null) {
                    cur.next = null;
                }

                count = 0;
                cur = next;
                ListNode head2 = next;

                /**
                 * 切分step个节点
                 */
                while (++count < step && cur != null) {
                    cur = cur.next;
                }
                next = cur == null ? null : cur.next;
                if (cur != null) {
                    cur.next = null;
                }

                tail.next = merge(head1, head2);

                while (tail.next != null) {
                    tail = tail.next;
                }

                cur = next;
            }
        }

        return newHead.next;

    }

    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        return mergeSort(head);
    }

    public static void main(String[] args) {
        ListNode n1 = new ListNode(4);
        ListNode n2 = new ListNode(2);
        ListNode n3 = new ListNode(1);
        ListNode n4 = new ListNode(3);

        n1.next = n2;
        n2.next = n3;
        n3.next = n4;

        ListNode node = new Solution().sortList(n1);
        while (node != null) {
            System.out.println(node.val);
            node = node.next;
        }
    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode() {
    }

    ListNode(int val) {
        this.val = val;
    }

    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}