[leetcode 23] Merge k Sorted Lists

最新推荐文章于 2025-09-11 19:19:55 发布

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最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： [LeetCode 刷题] 文章标签： leetcode Binary Search

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本文介绍了一种高效的算法来合并K个已排序的链表为一个排序链表。通过递归地将链表两两合并，最终实现整体排序。该方法在实际应用中能够有效地处理大规模数据排序问题。

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        int n = lists.size();
        if (n==0) {
            return NULL;
        }
        vector<ListNode *> cur;
        vector<ListNode *> res(lists);
        while (res.size()>1) {
            n = res.size()/2;
            for (int i = 0; i < n; i++) {
                cur.push_back(mergeTwoLists(res[2*i], res[2*i+1]));
            }
            if (res.size()%2) {
                cur.push_back(res[2*n]);
            }
            res = cur;
            cur.clear();
        }
        return res[0];
    }
    
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode *head = new ListNode(-1);
        ListNode *dummy = head;
        if (!l1) {
            return l2;
        } else if (!l2) {
            return l1;
        } else {
            while (l1 != NULL && l2 != NULL) {
                if (l1->val < l2->val) {
                    head->next = l1;
                    head = head->next;
                    l1 = l1->next;
                } else {
                    head->next = l2;
                    head = head->next;
                    l2 = l2->next;
                }
            }
            if (l1 == NULL) {
                head->next = l2;
            } else {
                head->next = l1;
            }
        }
        return dummy->next;
        
    }
};