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【算法题】合并K个升序链表

程序员漫画编程 2022-03-11 阅读 87
链表算法

给你一个链表数组,每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中,返回合并后的链表。

示例 1:

输入:lists = [[1,4,5],[1,3,4],[2,6]]
输出:[1,1,2,3,4,4,5,6]
解释:链表数组如下:
[
  1->4->5,
  1->3->4,
  2->6
]
将它们合并到一个有序链表中得到。
1->1->2->3->4->4->5->6
示例 2:

输入:lists = []
输出:[]
示例 3:

输入:lists = [[]]
输出:[]
 

提示:

k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i] 按 升序 排列
lists[i].length 的总和不超过 10^4

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) { 
        int k = lists.length;
        ListNode dummyHead = new ListNode(0);
        ListNode tail = dummyHead;
        while (true) {
            ListNode minNode = null;
            int minPointer = -1;
            for (int i = 0; i < k; i++) {
                if (lists[i] == null) {
                    continue;
                }
                if (minNode == null || lists[i].val < minNode.val) {
                    minNode = lists[i];
                    minPointer = i;
                }
            }
            if (minPointer == -1) {
                break;
            }
            tail.next = minNode;
            tail = tail.next;
            lists[minPointer] = lists[minPointer].next;
        }
        return dummyHead.next;
    }
}
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