leetcode23. 合并K个排序链表

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-lists

2. 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、将k个链表配对并将同一对中的链表合并;
2、重复这一过程，直到我们得到了终的有序链表。
*/

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: xidoublestar
method: 顺序合并
type: C
date: 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 == )
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: xidoublestar
method: 分治合并
type: C
date: 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 == )
return NULL;
return merge(lists, , 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)