一、问题描述
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
Example:
Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6
二、解题思路
这道题是合并K个list,本质上还是合并两个list(LeetCode:21. Merge Two Sorted Lists)。所以,每次取出来两个list进行合并即可,直到最终成为一个list
三、代码实现
#include <list>
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
if(lists.empty()) return NULL;
while (lists.size()>1){
int len = lists.size();
for (int i = 0; i < len/2; i++) {
//首、尾两个list合并(本质上就是Merge Two List)
lists[i] = mergeTwoLists(lists[i], lists[len - i - 1]);
}
//执行完一次for循环,list长度就减半。
//原来是3个,现在变成2;原来是4个,现在变成1个。
lists.resize((len + 1)/2);
}
//合并到最后lists中就剩下一个
return lists[0];
}
private:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if(l1 == NULL && l2 == NULL) return NULL;
else if(l1 != NULL && l2 == NULL) return l1;
else if(l1 == NULL && l2 != NULL) return l2;
else {
ListNode* preHead = new ListNode(0);
//preHead 需要移动,因此使用resultHead来先保存下
ListNode* resultHead = preHead;
while( l1!=NULL && l2!=NULL){
//同时不为空,那就比较。然后将小的添加到preHead的尾部
if(l1->val < l2->val){
preHead->next = l1;
preHead = preHead->next;
l1 = l1->next;
}else{
preHead->next = l2;
preHead = preHead->next;
l2 = l2->next;
}
//while结束,说明有一个为空了,那就把不为空的直接加上
if(l1!=NULL) preHead->next = l1;
if(l2!=NULL) preHead->next = l2;
}
return resultHead->next;
}
}
};