189 字
1 分钟
23. Merge k Sorted Lists
Problem
Soluitions
/** * 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) {
if(lists.length == 0) { return null; } // if(lists.length == 1) { // return lists[0]; // } for(int i = 1; i < lists.length; i++) { lists[i] = merge2Lists(lists[i-1], lists[i]); } return lists[lists.length - 1]; } public ListNode merge2Lists(ListNode l1, ListNode l2) { ListNode newList = new ListNode(); ListNode curr = newList; while(l1 != null && l2 != null) { if(l1.val < l2.val) { curr.next = l1; l1 = l1.next; } else { curr.next = l2; l2 = l2.next; } curr = curr.next; } if(l1 == null) { curr.next = l2; } if(l2 == null) { curr.next = l1; } return newList.next; }}暴力解决
- 解决合并2个顺序链表
- 循环调用直到合并完成整个list的有序链表
Reference
23. Merge k Sorted Lists
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