链表系列一＞合并 k 个升序链表

题目：

链接: link

方法一解析：

代码：

public ListNode mergeKLists(ListNode[] lists) {//建立小根堆PriorityQueue<ListNode> heap = new PriorityQueue<>((v1,v2)-> v1.val - v2.val);//把所有头节点放入小根堆中for(ListNode l : lists){if(l != null){heap.offer(l);}}ListNode ret = new ListNode(0);ListNode prev = ret;while(!heap.isEmpty()){ListNode t = heprev.next = t;prev.next = t;prev = t;if(t.next != null)heap.offer(t.next);}return ret.next;}

方法二解析：

代码：

//递归写法:public ListNode mergeKLists(ListNode[] lists) {return merge(lists,0,lists.length-1);}public ListNode merge(ListNode[] lists,int left, int right){if(left > right)return null;if(left == right)return lists[left];    //1.平分数组    int mid = (left + right) / 2;//[left,mid][mid+1,right]//2.递归处理ListNode l1 = merge(lists,left,mid);ListNode l2 = merge(lists,mid+1,right);//3.合并两个有序链表return megeToList(l1,l2);}private ListNode megeToList(ListNode l1, ListNode l2){if(l1 == null) return l2;if(l2 == null) return l1;ListNode ret = new ListNode(0);ListNode prev = ret;ListNode cur1 = l1, cur2 = l2;while(cur1 != null && cur2 != null){if(cur1.val < cur2.val){prev.next = cur1;prev = cur1;cur1 = cur1.next;}else{prev.next = cur2;prev = cur2;cur2 = cur2.next;}}if(cur1 != null) prev.next = cur1;if(cur2 != null) prev.next = cur2;return ret.next;}