这道题耗时40ms,思路是联想到快慢指针,vector有几个元素,我也有几个头指针,若A的头指针最小,就放入结果指针head的next中,则A就比别的指针快一步,移到下一位next。问题就是判断如何最小,可以自己写,也可以使用priority_queue这样有序的数据结构。感谢博客最下面的链接,我也学到了如何自定义比较器。代码如下:
struct cmp {
bool operator()(ListNode *&a, ListNode *&b) const {
return a->val > b->val;
}
};
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
if (lists.size() == 0) return nullptr;
priority_queue<ListNode *, vector<ListNode *>, cmp> q;
for (int i = 0; i < lists.size(); ++i) {
if (lists[i] != NULL) {
q.push(lists[i]);
}
}
ListNode* head = new ListNode(0);
ListNode* tail = head;
while (!q.empty()) {
tail->next = q.top();
q.pop();
tail = tail->next;
if (tail->next != NULL) {
q.push(tail->next);
}
}
return head->next;
}
};
最快实现代码如下,运用两个链表合并的方式:
class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {
if (lists.empty())
return NULL;
int len = lists.size();
while (len > 1) {
for (int i = 0; i < len / 2; ++i) {
lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);
}
len = (len + 1) / 2;
}
return lists[0];
}
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
ListNode *head = new ListNode(0);
ListNode *p = head;
while (l1 && l2) {
if (l1->val < l2->val) {
p->next = l1;
p = l1;
l1 = l1->next;
}
else {
p->next = l2;
p = l2;
l2 = l2->next;
}
}
if (l1)
p->next = l1;
else if (l2)
p->next = l2;
return head->next;
}
};