[Leetcode]Sort List

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

链表排序~可以用merge sort~找到链表的中点，然后对左右进行递归，最后再把两段lists合并~Merge Sort的时间复杂度为O(nlogn)~

class Solution:
    # @param head, a ListNode
    # @return a ListNode
    def sortList(self, head):
        if head is None or head.next is None: return head
        slow, fast = head, head
        while fast.next and fast.next.next:
            slow, fast = slow.next, fast.next.next
        head1, head2 = head, slow.next
        slow.next = None
        head1 = self.sortList(head1)
        head2 = self.sortList(head2)
        return self.mergeSort(head1, head2)
        
    def mergeSort(self, head1, head2):
        newHead = ListNode(0)
        cur = newHead
        while head1 and head2:
            if head1.val < head2.val:
                cur.next = head1
                head1 = head1.next
            else:
                cur.next = head2
                head2 = head2.next
            cur = cur.next
        if head1:
            cur.next = head1
        if head2:
            cur.next = head2
        return newHead.next