leetcode解题之23.Merge k Sorted Lists Java版本（合并k个有序的链表）

23. Merge k Sorted Lists

public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0)
return null;
return MSort(lists, 0, lists.length - 1);
}

public ListNode MSort(ListNode[] lists, int low, int high) {
if (low < high) {
int mid = (low + high) / 2;
ListNode leftlist = MSort(lists, low, mid);
ListNode rightlist = MSort(lists, mid + 1, high);
return mergeTwoLists(leftlist, rightlist);
}
// 如果相等，只有一个元素，返回即可
return lists[low];
}

// 递归合并链表
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode res = null;
if (l1 == null)
return l2;
if (l2 == null)
return l1;
if (l1.val <= l2.val) {
res = l1;
l1.next = mergeTwoLists(l1.next, l2);
} else {
res = l2;
l2.next = mergeTwoLists(l1, l2.next);
}
return res;
}

public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0)
return null;
// PriorityQueue 是堆，默认小顶堆
PriorityQueue<ListNode> min = new PriorityQueue<ListNode>(11, new Comparator<ListNode>() {
@Override
public int compare(ListNode o1, ListNode o2) {
return o1.val - o2.val;
}
});
// 加入所有链表的第一个结点，非空
for (ListNode node : lists)
if (node != null)
min.offer(node);
ListNode head = new ListNode(0);
ListNode cur = head;
while (!min.isEmpty()) {
ListNode temp = min.poll();
cur.next = temp;
cur = cur.next;
// 边取边加入
if (temp.next != null)
min.offer(temp.next);
}
// 注意断链
cur.next = null;
return head.next;
}