LeetCode 23 合并K个排序链表

最新推荐文章于 2022-07-09 15:33:33 发布

TIMELIMITE

最新推荐文章于 2022-07-09 15:33:33 发布

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本文介绍了一种使用Java实现的链表合并算法，通过递归和迭代的方式将多个链表合并为一个有序链表。文章详细解释了算法的实现过程，包括如何处理不同长度的链表以及如何使用优先队列进行优化。

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// 解法一: 采用归并的思路,两两合并
// 解法二:小子只是想熟悉Java优先队列使用方法,但提交报错


class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null)
            return null;
        if (lists.length == 0)
            return null;
        if (lists.length == 1)
            return lists[0];
        if (lists.length == 2)
            return mergeTwoList(lists[0], lists[1]);

        int l = 0;
        int r = lists.length - 1;
        int mid = (l + r) / 2;
        ListNode left = mergePartList(lists, l, mid);
        ListNode right = mergePartList(lists, mid + 1, r);
        return mergeTwoList(left, right);
    }

    public ListNode mergePartList(ListNode[] lists, int l, int r) {

        if (l > r)
            return null;
        if (l == r)
            return lists[l];

        int mid = (l + r) / 2;
        ListNode left = mergePartList(lists, l, mid);
        ListNode right = mergePartList(lists, mid + 1, r);
        return mergeTwoList(left, right);

    }

    public ListNode mergeTwoList(ListNode l1, ListNode l2) {
        ListNode res = null;

        if (l1 == null)
            return l2;
        if (l2 == null)
            return l1;

        if (l1.val < l2.val) {
            res = l1;
            l1 = l1.next;
        } else {
            res = l2;
            l2 = l2.next;
        }

        ListNode head = res;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                res.next = l1;
                l1 = l1.next;
                res = res.next;
            } else {
                res.next = l2;
                l2 = l2.next;
                res = res.next;
            }
        }

        while (l1 != null) {
            res.next = l1;
            l1 = l1.next;
            res = res.next;
        }
        while (l2 != null) {
            res.next = l2;
            l2 = l2.next;
            res = res.next;
        }
        res.next = null;
        return head;
    }

    // public ListNode mergeKLists(ListNode[] lists) {
    //     if (lists == null || lists.length == 0)
    //         return null;
    //     if (lists.length == 1)
    //         return lists[0];
    //     PriorityQueue<ListNode> que = new PriorityQueue<>(new Comparator<ListNode>() {
    //         public int compare(ListNode o1, ListNode o2) {
    //             return o1.val - o2.val;
    //         }
    //     });
    //     ListNode head = null;
    //     for (ListNode list : lists) {
    //         while (list != null) {
    //             que.add(list);
    //             list = list.next;
    //         }
    //     }
    //     ListNode res = new ListNode(-1);
    //     head = res;
    //     while (!que.isEmpty()) {
    //         head.next = que.poll();
    //         head = head.next;
    //     }
    //     if (head != null)
    //         head.next = null;
    //     return res.next;
    // }
}