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23. Merge k Sorted Lists
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
归并排序o(nlgk);
/** heap * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode mergeKLists(ListNode[] lists) { if (lists == null || lists.length == 0) return null; else if (lists.length == 1) return lists[0]; return divide(lists, 0, lists.length - 1); } private ListNode divide(ListNode[] lists, int begin, int end) { if (end <= begin) return lists[begin]; int middle = (begin + end) / 2; ListNode left = divide(lists, begin, middle); ListNode right = divide(lists, middle + 1, end); return mergeTwoList(left, right); } public ListNode mergeTwoList(ListNode l1, ListNode l2){ if(l1 == null) return l2; if(l2 == null) return l1; if(l1.val < l2.val){ l1.next = mergeTwoList(l1.next , l2); return l1; } else{ l2.next = mergeTwoList(l1 ,l2.next); return l2; } } }
法二 : 优先队列
维护一个长为n的队列 o(n *lgk)
public ListNode mergeKLists(ListNode[] lists) { PriorityQueue<ListNode> queue = new PriorityQueue<ListNode>(new Comparator<ListNode>(){ public int compare(ListNode l1, ListNode l2){ return l1.val - l2.val; } }); for(ListNode n: lists){ if(n != null) queue.offer(n); } ListNode dummy = new ListNode(0); ListNode tail = dummy; while(!queue.isEmpty()){ ListNode ln = queue.poll(); tail.next = ln; tail = tail.next; if(ln != null && ln.next != null) queue.add(ln.next); } return dummy.next; }
23. Merge k Sorted Lists
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