Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
合并K个有序链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* merge(vector<ListNode*>&lists, int start, int end){
if(start==end)return lists[start];
int mid=(start+end)/2;
ListNode*p1=merge(lists, start, mid);
ListNode*p2=merge(lists, mid+1, end);
ListNode*head=NULL, *p=NULL;
while(p1&&p2){
if(p1->val<p2->val){
if(p)p->next=p1;
p=p1;
p1=p1->next;
}
else{
if(p)p->next=p2;
p=p2;
p2=p2->next;
}
if(head==NULL)head=p;
}
if(p1){
if(p)p->next=p1;
else{p=p1;head=p;}
}
if(p2){
if(p)p->next=p2;
else{p=p2;head=p;}
}
return head;
}
ListNode *mergeKLists(vector<ListNode *> &lists) {
int size=lists.size();
if(size==0) return NULL;
if(size==1) return lists[0];
return merge(lists, 0, size-1);
}
};LeetCode: Merge k Sorted Lists [022],布布扣,bubuko.com
LeetCode: Merge k Sorted Lists [022]
原文地址:http://blog.csdn.net/harryhuang1990/article/details/26007585