题目描述:

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

解题思路:

分治方法。将K个List不断地分解为前半部分和后半部分。分别进行两个List的合并。最后将合并的结果合并起来。

代码如下:

/**
* 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.length == 0)
return null;
return divide(lists, 0, lists.length - 1);
} public ListNode divide(ListNode[] l1, int left, int right) {
if(left < right){
int mid = (left + right) / 2;
return mergeTwoLists(divide(l1, left, mid), divide(l1, mid + 1, right));
}
return l1[left];
} public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode list = new ListNode(0);
ListNode tmp = list;
while (l1 != null || l2 != null) {
if (l1 == null) {
tmp.next = new ListNode(l2.val);
l2 = l2.next;
} else if (l2 == null) {
tmp.next = new ListNode(l1.val);
l1 = l1.next;
} else {
if (l1.val < l2.val) {
tmp.next = new ListNode(l1.val);
l1 = l1.next;
} else {
tmp.next = new ListNode(l2.val);
l2 = l2.next;
}
}
tmp = tmp.next;
}
return list.next;
}
}

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