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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
以下程序实现了这一功能,请你填补空白处内容:
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) {
if (lists.length == 0)
return null;
return merge(lists, 0, lists.length - 1);
}
public ListNode merge(ListNode[] lists, int low, int high) {
if (high - low == 0)
return lists[low];
else if (high - low == 1)
return mergeTwoLists(lists[low], lists[high]);
else {
int mid = (low + high) / 2;
_____________________________;
return mergeTwoLists(tmp1, tmp2);
}
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode();
ListNode p = head;
while (l1 != null && l2 != null) {
if (l1.val > l2.val) {
p.next = l2;
l2 = l2.next;
p = p.next;
} else {
p.next = l1;
l1 = l1.next;
p = p.next;
}
}
if (l1 != null)
p.next = l1;
if (l2 != null)
p.next = l2;
return head.next;
}
}
Python每日一练 https://edu.csdn.net/practice/23630877 中的主程序中合并的是列表,而不是链表
l = LinkList()
list1 = [[1, 4, 5], [1, 3, 4], [2, 6]]
s = Solution()
print(l.convert_list(s.mergeKLists(list1)))
中间插入链表初始化:
for i,lst in enumerate(list1):
list1[i] = l.initList(lst)