題目描述
合併K個排序鏈表
合併 k 個排序鏈表,返回合併後的排序鏈表。請分析和描述算法的複雜度。
示例:
輸入:
[
1->4->5,
1->3->4,
2->6
]
輸出: 1->1->2->3->4->4->5->6
解答思路
由於題目中給了我們k個鏈表,所以可以使用分治的思想來進行解決。具體就是將k個鏈表的範圍不斷進行二分,在遞歸的base條件中再將兩條鏈表合併爲一條有序的鏈表,跟歸併排序的思想類似。
具體思路如下
代碼
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) {
return null;
}
return mergeSort(lists, 0, lists.length - 1);
}
public ListNode mergeSort(ListNode[] lists, int left, int right) {
if (left >= right) {
return lists[left];
}
int mid = left + ((right - left) >> 1);
ListNode l1 = mergeSort(lists, left, mid);
ListNode l2 = mergeSort(lists, mid + 1, right);
return merge(l1, l2);
}
public ListNode merge(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode cur = dummyHead;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
cur.next = l1;
cur = cur.next;
l1 = l1.next;
} else {
cur.next = l2;
cur = cur.next;
l2 = l2.next;
}
}
if (l1 != null) {
cur.next = l1;
}
if (l2 != null) {
cur.next = l2;
}
return dummyHead.next;
}
}