23.合并k个有序表

Merge k Sorted Lists

问题描述：

知识补充：

lists.erase(lists.begin()+i);//删除第i个位置的元素，注意此时vector动态改变，即此时i+1位置的元素变为现在的i位置
lists.erase(lists.begin()+i,lists.begin()+j);//删除区间[i,j)的元素

priority_queue<Type, Container, Functional> priQ;//Type 为数据类型， Container 为保存数据的容器，Functional 为元素比较方式。
priQ.push();//元素加入队列
priQ.top();//返回队列头部数据
priQ.pop();//队列头部数据出队
priQ.empty();//判断队列是否为空
priQ.size();//返回队列中数据的个数

测试代码：

ListNode* mergeKLists(vector<ListNode*>& lists) {
int i=0,pos = 0;
if(lists.empty())
return NULL;
while(i<lists.size())
{
if(!lists[i])
{
lists.erase(lists.begin()+i);
continue;
}
i++;
}
ListNode *p=lists[0];
ListNode dummy(INT_MIN);
ListNode *tail = &dummy;

while(!lists.empty())
{
i = 0;
p = lists[0];
pos = 0;
while(i<lists.size())
{

if(lists[i]->val<p->val)
{
p = lists[i];
pos = i;
}
i++;
}
tail->next = p;
tail = tail->next;
if(p->next)
{
p = p->next;
lists[pos] = p;
}else
{
lists.erase(lists.begin()+pos);
}
}

return dummy.next;

}

性能：

参考答案：

class Solution {
public:
// Need to use > because we want priority queue to return the min priority first
struct priQueueCmp {
bool operator() (const ListNode* l, const ListNode* r) {
return l->val > r->val;
}
};

ListNode* mergeKLists(vector<ListNode*>& lists) {
// Using priority queue to pick out min of the lists at each step.
// Pri queue picks out the min in const time, but insertion and deletion takes
// log time
priority_queue<ListNode*, vector<ListNode*>, priQueueCmp> priQ;

ListNode* head = NULL;
ListNode* lastNode = NULL;
// load up
for(int i = 0; i < lists.size(); ++i) {
if(lists[i] != NULL)
priQ.push(lists[i]);
}

if(priQ.empty())
return head;

head = priQ.top();
lastNode = head;
priQ.pop();
if(head->next != NULL)
priQ.push(head->next);

while(!priQ.empty())
{
ListNode* minNode = priQ.top();
lastNode->next = minNode;
lastNode = minNode;
priQ.pop();
if(minNode->next != NULL)
priQ.push(minNode->next);

}

return head;
}
};

性能：