此题使用归并思想解题,步骤类似于归并排序方法
题目:
合并 k 个排序链表,返回合并后的排序链表。请分析和描述算法的复杂度。
示例:
输入:
[
1->4->5,
1->3->4,
2->6
]
输出: 1->1->2->3->4->4->5->6
代码如下所示:
/**
* 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 nullptr;
// 开始分解数组
return separation(lists, 0, lists.size() - 1);
}
private:
ListNode* merge(ListNode* l1, ListNode* l2) {
if (l1 == nullptr) return l2;
else if (l2 == nullptr) return l1;
ListNode* retList = nullptr;
ListNode* p = l1, * q = l2;
p->val > q->val ? (retList = new ListNode(q->val), q = q->next) :
(retList = new ListNode(p->val), p = p->next);
ListNode* cur = retList;
while (p != nullptr && q != nullptr) {
if (p->val > q->val) {
cur->next = new ListNode(q->val);
q = q->next;
}
else {
cur->next = new ListNode(p->val);
p = p->next;
}
cur = cur->next;
}
while (p != nullptr) {
cur->next = new ListNode(p->val);
cur = cur->next;
p = p->next;
}
while (q != nullptr) {
cur->next = new ListNode(q->val);
cur = cur->next;
q = q->next;
}
return retList;
}
ListNode* separation(vector<ListNode*>& lists, int left, int right) {
if (left >= right) return lists[left];
int mid = (left + right) / 2;
// 递归分解左边与右边
ListNode* p = separation(lists, left, mid);
ListNode* q = separation(lists, mid + 1, right);
// 开始合并(两个链表的排序合并)
return merge(p, q);
}
};