sort-list 归并排序链表
题目：
在O(n log n)的时间内使用常数级空间复杂度对链表进行排序。
Sort a linked list in O(n log n) time using constant space complexity.
示例;
输入：{3,2,4} 输出：{2,3,4}
代码：
1 /** 2 * struct ListNode { 3 * int val; 4 * struct ListNode *next; 5 * }; 6 */ 7 8 class Solution { 9 public: 10 /** 11 * 12 * @param head ListNode类 13 * @return ListNode类 14 */ 15 ListNode* FindMiddleNode(ListNode* head) { 16 ListNode* fast = head; //利用快慢指针找到当前中间节点 17 ListNode* slow = head; 18 while(fast->next != NULL && fast->next->next != NULL){ 19 fast = fast->next->next; 20 slow = slow->next; 21 } 22 return slow; 23 } 24 ListNode* MergeList(ListNode* l1, ListNode* l2) { 25 if(l1 == NULL) 26 return l2; 27 if(l2 == NULL) 28 return l1; 29 ListNode* result; //结果 30 ListNode* newhead; //头节点 31 auto first = l1->next; //表示归并排序中的两个分区 32 auto second = l2->next; 33 if(l1->val > l2->val){ //l1较小则为头节点，反之则头节点为l2 34 result = newhead = l2; 35 first = l1; 36 } 37 else{ 38 result = newhead = l1; 39 second = l2; 40 } 41 while(first != NULL || second != NULL){ 42 if(second == NULL){ //当第二块为空 43 result->next = first; break; 44 } 45 else if(first == NULL){ //当第一块为空 46 result->next = second; break; 47 } 48 // 归并排序，较小的先插入result的尾部，由于这里在sort函数中有将右末尾结点设为NULL所以会停止循环 49 if(first->val > second->val){ 50 result->next = second; 51 second = second->next; 52 result = result->next; 53 } 54 else{ 55 result->next = first; 56 first = first->next; 57 result = result->next; 58 } 59 } 60 return newhead; 61 } 62 ListNode* sortList(ListNode* head) { 63 if(head == NULL || head->next == NULL) 64 return head; 65 ListNode* mid = FindMiddleNode(head); 66 ListNode* midnext = mid->next; 67 mid->next = NULL; //将链表划分 68 return MergeList(sortList(head), sortList(midnext)); 69 } 70 };
我的笔记：
由于本函数的要求时间复杂度为O(nlogn)，故使用归并排序法。
leetcode官网详解：https://leetcode-cn.com/problems/sort-list/solution/148-pai-xu-lian-biao-bottom-to-up-o1-kong-jian-by-
