LeetCode算法题148：排序链表解析

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。

示例 1:

输入: 4->2->1->3
输出: 1->2->3->4

示例 2:

输入: -1->5->3->4->0
输出: -1->0->3->4->5

这个题需要用到归并排序，归并排序有两个过程，一个是把序列分开，一个是把序列合并，这里需要用到两个函数，使用递归来进行编程，每一次递归都是分开序列和合并序列一起操作。对于归并排序不清楚的可以维基百科搜索一下，看一下代码，就会清楚很多。

C++源代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* sortList(ListNode* head) {
        if(!head || !head->next) return head;
        ListNode *slow = head, *fast = head, *pre = head;
        while(fast && fast->next){
            pre = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        pre->next = NULL;
        return merge(sortList(head), sortList(slow));
    }
    ListNode* merge(ListNode* left, ListNode* right){
        ListNode *dummy = new ListNode(-1);
        ListNode *cur = dummy;
        while(left && right){
            if(left->val < right->val){
                cur->next = left;
                left = left->next;
            }
            else{
                cur->next = right;
                right = right->next;
            }
            cur = cur->next;
        }
        if(left) cur->next = left;
        if(right) cur->next = right;
        return dummy->next;
    }
};

python3源代码：

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        if head==None or head.next==None:
            return head
        slow = fast = pre = head
        while fast and fast.next:
            pre = slow
            slow = slow.next
            fast = fast.next.next
        pre.next = None
        return self.merge(self.sortList(head), self.sortList(slow))
    def merge(self, left, right):
        if left==None:
            return right
        if right==None:
            return left
        if left.val < right.val:
            left.next = self.merge(left.next, right)
            return left
        else:
            right.next = self.merge(left, right.next)
            return right