Screenshot from 2016-01-23 15:58:48.png
My code:
/**
* 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 + 1 < 2)
return lists[begin];
int middle = (begin + end) / 2;
ListNode left = divide(lists, begin, middle);
ListNode right = divide(lists, middle + 1, end);
return mergeTwoLists(left, right);
}
private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null)
return l2;
else if (l2 == null)
return l1;
ListNode dummy = new ListNode(-1);
ListNode temp = dummy;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
temp.next = l1;
l1 = l1.next;
}
else {
temp.next = l2;
l2 = l2.next;
}
temp = temp.next;
}
temp.next = (l1 == null) ? l2 : l1;
return dummy.next;
}
}
我就是采用了merge sort 的思想，把这些lists先divide，到只剩下两个的时候，再merge一下。算了下复杂度，其实挺高的。每个list都需要重复访问多次把。
比如0, 1, 2, 3 条list
那么 0 ,1 merge， 2 和3 merge
然后 0,1组成的综合体与2,3组成的综合体再merge
这个时候又需要访问下0,1,2,3。 有重复。时间复杂度应该比较高。虽然看测试下来，效果还不错。
后来看了一个新的解法，复杂度要低很多。用priority queue 来做。
My code:
import java.util.Comparator;
import java.util.PriorityQueue;
/**
* 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;
PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>(lists.length, new Comparator<ListNode>() {
public int compare(ListNode o1, ListNode o2) {
if (o1.val < o2.val)
return -1;
else if (o1.val == o2.val)
return 0;
else
return 1;
}
});
for (int i = 0; i < lists.length; i++) {
if (lists[i] != null)
pq.add(lists[i]);
}
ListNode dummy = new ListNode(-1);
ListNode scanner = dummy;
while (pq.size() > 0) {
ListNode temp = pq.poll();
scanner.next = temp;
scanner = scanner.next;
if (temp.next != null)
pq.add(temp.next);
}
return dummy.next;
}
}
具体看下代码就懂了。复杂度是 log(k) * n
k = number of lists
n = number of all nodes
注意下这道题目用到了 PrioirtyQueue 这个数据结构，并且可以自带Comparator进去。
记得上次使用到的数据结构， TreeSet, 其实就是Binary search tree
参考网站：
http://www.programcreek.com/2013/02/leetcode-merge-k-sorted-lists-java/
Anyway, Good luck, Richardo!
My code:
import java.util.PriorityQueue;
import java.util.Comparator;
/**
* 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;
}
PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>(8, new Comparator<ListNode>() {
public int compare(ListNode o1, ListNode o2) {
return o1.val - o2.val;
}
});
for (int i = 0; i < lists.length; i++) {
if (lists[i] != null) {
pq.offer(lists[i]);
}
}
ListNode dummy = new ListNode(-1);
ListNode pre = dummy;
while (!pq.isEmpty()) {
ListNode curr = pq.poll();
if (curr.next != null) {
pq.offer(curr.next);
}
pre.next = curr;
pre = curr;
}
return dummy.next;
}
public static void main(String[] args) {
Solution test = new Solution();
System.out.println("111");
}
}
how to initialize a PriorityQueue object
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(8, new Comparator<Integer>() {
public int compare(Integer i1, Integer i2) {
return i1.compareTo(i2);
}
});
觉得用pq 来做这道题目是最好的。
Anyway, Good luck, Richardo! -- 09/14/2016
