【Leetcode】Merge K Sorted Linked List

【题目】

Merge k sorted Linked Lists and return it as one sorted lists. Analyze and describe its complexity .

【思路】

leetcode clean book

【代码】

private static final Comparator<ListNode> listComparator = new Comparator<ListNode>(){public int compare(ListNode x, ListNode y){return x.val - y.val;}};public static ListNode mergeK(List<ListNode> list){if(list.isEmpty()) return null;Queue<ListNode> queue = new PriorityQueue<>(list.size(),listComparator);for(ListNode node : list){if(node!=null){queue.add(node);}}ListNode dummy = new ListNode(0);ListNode p = dummy;while(!queue.isEmpty()){ListNode node = queue.poll();p.next = node;p=p.next;if(node.next!=null){queue.add(node.next);}}return dummy.next;}

public static ListNode mergeK2(List<ListNode> list){if (list == null) return null;int end = list.size() - 1;while(end > 0){int start = 0;while(start < end){list.set(start, merge2List(list.get(start),list.get(end)));start++;end--;}}return list.get(0);}public static ListNode merge2List(ListNode l1, ListNode l2){ListNode dummy = new ListNode(0);ListNode cur = dummy; ListNode p = l1, q = l2; while(p != null && q != null){ if(p.val > q.val ){ cur.next = q; q = q.next; }else{ cur.next = p; p = p.next; } cur = cur.next; } if(p!=null) cur.next = p; if(q!=null) cur.next = q;  return dummy.next;}