leetcode23.合并K個(gè)排序鏈表

1. 題目描述

合并 k 個(gè)排序鏈表，返回合并后的排序鏈表。請分析和描述算法的復(fù)雜度。
示例:
輸入:
[
  1->4->5,
  1->3->4,
  2->6
]
輸出: 1->1->2->3->4->4->5->6
來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/merge-k-sorted-lists

2. 解題思路

/*
解題思路：
解法一、順序合并
1、lists[0]與lists[1]合并，結(jié)果與lists[2]合并...結(jié)果與lists[listsSize-1]合并
解法二、分治合并
1、lists[0]與lists[1]合并,lists[2]與lists[3]合并,然后將合并的結(jié)果繼續(xù)合并。
*/

3. 測試結(jié)果

解法一、順序合并

解法二、分治合并

4. 順序合并

/*
title: leetcode23. 合并K個(gè)排序鏈表
author: xidoublestar
method: 順序合并
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 == 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個(gè)排序鏈表
author: xidoublestar
method: 順序合并
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 == 0)
        return NULL;
    return merge(lists, 0, listsSize - 1);
}

6. 復(fù)雜度分析

解法一、順序合并
時(shí)間復(fù)雜度：O(n*n)
空間復(fù)雜度：O(1)
解法二、分治合并
時(shí)間復(fù)雜度：O(nlogn)
空間復(fù)雜度：O(1)