Sort a linked list in O(n log n) time using constant space complexity.
单链表排序----快排 & 归并排序
（1）归并排序
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* sortList(ListNode* head) { 12 if (head == nullptr || head->next == nullptr) 13 return head; 14 ListNode* slow = head; 15 ListNode* fast = head; 16 ListNode* mid = nullptr; 17 while (fast&&fast->next) 18 { 19 mid = slow; 20 slow = slow->next; 21 fast = fast->next->next; 22 } 23 mid->next = nullptr; 24 return mergeHelp(sortList(head), sortList(slow)); 25 } 26 ListNode* mergeHelp(ListNode* low, ListNode* high) 27 { 28 ListNode* head = new ListNode(-1); 29 ListNode* cur = head; 30 while (low&&high) 31 { 32 if (low->val < high->val) 33 { 34 cur->next = low; 35 low = low->next; 36 } 37 else 38 { 39 cur->next = high; 40 high = high->next; 41 } 42 cur = cur->next; 43 } 44 cur->next = low ? low : high; 45 return head->next; 46 } 47 };
（2）快速排序
①.使第一个节点为中心点；
②.创建2个指针(first，second)，first指向头结点，second指向first的下一个节点；
③.second开始遍历,如果发现second的值比中心点的值小，则此时first=first.next，并且执行当前first的值和second的值交换，second遍历到链表尾即可；
④.把头结点的值和first的值执行交换。此时first节点为中心点,并且完成1轮快排
⑤.使用递归的方法即可完成排序检测是否合法
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* sortList(ListNode* head) { 12 if(head==nullptr) 13 return head; 14 quickSort(head,nullptr); 15 return head; 16 } 17 void quickSort(ListNode* begin,ListNode* end) 18 { 19 if(begin!=end) 20 { 21 ListNode* sep=getSeperator(begin,end); 22 quickSort(begin,sep); 23 quickSort(sep->next,end); 24 } 25 } 26 ListNode* getSeperator(ListNode* begin,ListNode* end) 27 { 28 ListNode* first=begin; 29 ListNode* second=begin->next; 30 int key=begin->val; 31 while(second!=end) 32 { 33 if(second->val<key) 34 { 35 first=first->next; 36 swap(first,second); 37 } 38 second=second->next; 39 } 40 swap(begin,first); 41 return first; 42 } 43 void swap(ListNode* a,ListNode* b) 44 { 45 int tmp=a->val; 46 a->val=b->val; 47 b->val=tmp; 48 } 49 };