1. 题目描述

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。示例:输入:[ 1->4->5, 1->3->4, 2->6]输出: 1->1->2->3->4->4->5->6来源：力扣（LeetCode）链接：https://leetcode-cn.com/problems/merge-k-sorted-lists2. 解题思路

/*解题思路：解法一、顺序合并1、lists[0]与lists[1]合并，结果与lists[2]合并...结果与lists[listsSize-1]合并解法二、分治合并1、lists[0]与lists[1]合并,lists[2]与lists[3]合并,然后将合并的结果继续合并。*/3. 测试结果

解法一、顺序合并

解法二、分治合并
4. 顺序合并

/*title: leetcode23. 合并K个排序链表author: xidoublestarmethod: 顺序合并type: Cdate: 2020-5-27*/struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) { if (!l1) return l2; if (!l2) return l1; struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode)), * tail = head; while (l1 && l2) { if (l1->val < l2->val) { tail->next = l1; l1 = l1->next; } else { tail->next = l2; l2 = l2->next; } tail = tail->next; } if (l1) tail->next = l1; else if (l2) tail->next = l2; tail = head; head = head->next; free(tail); return head;}struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) { if (listsSize == 0) return NULL; struct ListNode* res = *lists; for (int i = 1; i < listsSize; i++) { if(lists[i] != NULL) res = mergeTwoLists(res, lists[i]); } return res;}5. 分治合并

/*title: leetcode23. 合并K个排序链表author: xidoublestarmethod: 顺序合并type: Cdate: 2020-5-27*/struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) { if ((!l1) || (!l2)) return l1 ? l1 : l2; struct ListNode head; head.next = NULL; struct ListNode* tail = &head; while (l1 && l2) { if (l1->val < l2->val) { tail->next = l1; l1 = l1->next; } else { tail->next = l2; l2 = l2->next; } tail = tail->next; } tail->next = l1 ? l1 : l2; return head.next;}struct ListNode* merge(struct ListNode** lists, int left, int right) { if (left == right) return lists[left]; if (left > right) return NULL; int mid = (left + right) >> 1; struct ListNode* p1 = merge(lists, left, mid); struct ListNode* p2 = merge(lists, mid + 1, right); return mergeTwoLists(p1, p2);}struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) { if (listsSize == 0) return NULL; return merge(lists, 0, listsSize - 1);}6. 复杂度分析

解法一、顺序合并
时间复杂度：O(n*n)
空间复杂度：O(1)
解法二、分治合并
时间复杂度：O(nlogn)
空间复杂度：O(1)