148 Sort List

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

Example：

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

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

解释下题目：

链表排序，时间复杂度要求nlogn，且空间复杂度是常量

1. 方法1

实际耗时：xxms

public ListNode sortList(ListNode head) {
    if (head == null || head.next == null) {
        return head;
    }
    ListNode fast = head;
    ListNode slow = head;
    ListNode pre = slow;
    while (fast != null && fast.next != null) {
        fast = fast.next.next;
        pre = slow;
        slow = slow.next;
    }
    // now slow is in the middle
    // !!! don't forget to cut this list
    pre.next = null;
    return mergeList(sortList(head), sortList(slow));
}

private ListNode mergeList(ListNode l1, ListNode l2) {
    ListNode res = new ListNode(0);
    ListNode cur = res;
    while (l1 != null && l2 != null) {
        if (l1.val < l2.val) {
            cur.next = l1;
            l1 = l1.next;
            cur = cur.next;
        } else {
            cur.next = l2;
            l2 = l2.next;
            cur = cur.next;
        }
    }
    if (l1 != null) {
        cur.next = l1;
    }
    if (l2 != null) {
        cur.next = l2;
    }
    return res.next;
}

  思路跟数组的排序也是一样的，找到中间，切断，然后一分为二继续，而且我感觉merge甚至比数组还简单。唯一需要注意的就是记得切断。
  而且题目说的是常量空间，其实递归的话应该就不算了，但是如果要nlogn确实也只有快排了。

时间复杂度O(nlogn)

空间复杂度O(1)

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