第二周Merge k Sorted Lists

最新推荐文章于 2024-04-03 21:24:25 发布

原创最新推荐文章于 2024-04-03 21:24:25 发布 · 213 阅读

0 ·

CC 4.0 BY-SA版权

本文深入解析了LeetCode上第23题“合并K个有序链表”的算法实现，采用分治策略将多个有序链表高效合并为一个。详细介绍了算法步骤，包括如何将链表分组、递归排序以及最终合并链表的过程。

Merge k Sorted Lists

本周讲的是分治，所以选的题目也是分治的问题
LeetCode上的23题：https://leetcode.com/problems/merge-k-sorted-lists/description/

Merge k Sorted Lists

题目

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

题解

分：当要排序的链表大于2时，将链表分成左右两个部分排序，若要排序的链表数为一时，拷贝该链表并返回
治：对两个已排序链表进行排序，返回包含两个链表元素的已排序链表

代码

class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        if (lists.size() == 0) return NULL;
        return _mergeKlists(lists, 0, lists.size() - 1);
    }

    ListNode* _mergeKlists(vector<ListNode*>& lists, int left, int right) {
        ListNode* newList = new ListNode(0);
        ListNode* cur = newList;

        if (left == right) {
            ListNode* p = lists[left];
            while (p != NULL) {
                cur->next = new ListNode(p->val);
                p = p->next;
                cur = cur->next;
            }
        }
        else {
            int mid = (left + right) / 2;
            ListNode* l = _mergeKlists(lists, left, mid);
            ListNode* r = _mergeKlists(lists, mid + 1, right);
            while (l != NULL && r != NULL) {
                if (l->val < r->val) {
                    cur->next = l;
                    l = l->next;
                }
                else {
                    cur->next = r;
                    r = r->next;
                }
                cur = cur->next;
            }
            if (l == NULL) {
                cur->next = r;
            }
            else {
                cur->next = l;
            }
        }

        cur = newList->next;
        delete newList;
        return cur;
    }
};