一、問題描述
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、解題思路
這道題是合併K個list,本質上還是合併兩個list(LeetCode:21. Merge Two Sorted Lists)。所以,每次取出來兩個list進行合併即可,直到最終成爲一個list
三、代碼實現
#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、尾兩個list合併(本質上就是Merge Two List)
lists[i] = mergeTwoLists(lists[i], lists[len - i - 1]);
}
//執行完一次for循環,list長度就減半。
//原來是3個,現在變成2;原來是4個,現在變成1個。
lists.resize((len + 1)/2);
}
//合併到最後lists中就剩下一個
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 需要移動,因此使用resultHead來先保存下
ListNode* resultHead = preHead;
while( l1!=NULL && l2!=NULL){
//同時不爲空,那就比較。然後將小的添加到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結束,說明有一個爲空了,那就把不爲空的直接加上
if(l1!=NULL) preHead->next = l1;
if(l2!=NULL) preHead->next = l2;
}
return resultHead->next;
}
}
};