題目:
Merge k sorted
linked lists and return it as one sorted list. Analyze and describe its complexity.
分析:
1. 用暴力的方法一個一個合併鏈表會超時,分析所需要的時間複雜度:
假設鏈表的長度爲n,每兩個鏈表合併所需要的時間爲n+n,k個鏈表一共需要合併k-1次,那麼公式可以寫爲:
2n+3n+4n+..+kn = O(n*k^2)
2. 使用歸併排序的思想,兩個兩個的進行合併。兩兩合併的代碼和Merge Two Sorted Lists那一題一樣。分析一下這個方法所需要的時間複雜度:
把k個鏈表兩兩合併,問題規模的變化過程:k --> k/2 --> k/4 -->..-->1 一共需要logk次。
每次合併時鏈表的規模爲 : n + 2n + 4n + 8n + .. + 2^(logk)n
所需要的時間複雜度爲: n*k + 2n*k/2 + 4n*k/4 + .. + 1*2^(logk)n = nk+nk+nk+..+nk = nklogk
Java代碼實現:
/**
* 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;
if(lists.length==1)
return lists[0];
int end = lists.length-1;
while(end>0)
{
int start = 0;
while(start<end)
{
lists[start] = merge2lists(lists[start], lists[end]);
start++;
end--;
}
}
return lists[0];
}
private ListNode merge2lists(ListNode l1, ListNode l2)
{
if(l1==null)
return l2;
if(l2==null)
return l1;
ListNode dummy = new ListNode(0);
ListNode node = dummy;
while(l1!=null && l2!=null)
{
if(l1.val<=l2.val)
{
node.next = l1;
l1 = l1.next;
}
else
{
node.next = l2;
l2 = l2.next;
}
node = node.next;
}
while(l1!=null)
{
node.next = l1;
node = node.next;
l1 = l1.next;
}
while(l2!=null)
{
node.next = l2;
node = node.next;
l2 = l2.next;
}
return dummy.next;
}
}