LeetCode 148. Sort List 單鏈表排序 Python Solution

此題目對應於LeetCode 148

題目要求：

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

用O(n log n) 的排序算法，很容易讓人想到用遞推去做。

思路就是將單鏈表一分爲二劃分成兩個子問題，然後遞歸調用算法，最後在做一個對兩個

有序的單鏈表的合併操作(O(n))。

按照主定理：

T(n) = 2T(n/2) + n

則 T(n) = n = f(n), 則 T(n) =  O(nlogn)

單鏈表一分爲二就直接從單鏈表中間劃分

下面附上python代碼：

class ListNode:  
    def __init__(self,x):  
        self.val=x
        self.next=None 
    def mergeTwoLists(self, l1, l2):#合併兩個有序的鏈表
        if not l1:
            return l2
        if not l2:
            return l1
        prenode = ListNode(1)
        p = prenode
        while l1 and l2:
            if l1.val <= l2.val:
                p.next = l1
                l1 = l1.next
            else:
                p.next = l2
                l2 = l2.next
            p = p.next
        if l1:
            p.next = l1
        if l2:
            p.next = l2
        return prenode.next
    
    def sortList(self, head):
        if not head or not head.next:
            return head
        prenode = head
        p1 = prenode
        p2 = prenode
        l = 0
        while p1:
            l += 1
            p1 = p1.next
        mid = l/2
        k = 0
        l1 = prenode
        while p2: # 將單鏈表從中間一分爲二
            if k >= mid-1:
                tmp = p2.next
                p2.next = None
                l2 = tmp
                break
            else:
                p2 = p2.next
                k += 1
        t1 = self.sortList(l1)
        t2 = self.sortList(l2)
        return self.mergeTwoLists(t1,t2)