Problem: 148. 排序链表
解析
参考合并两个有序链表,那么需要做的只是把链表递归分为两半,然后合并两半即可，即归并排序，
slow和fast指针的使用可以找到要分割的点。
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class Solution {
public:
ListNode *sortList(ListNode *head) { return sortList(head, nullptr); }
ListNode *sortList(ListNode *head, ListNode *tail) {
if (head == nullptr) {
return nullptr;
}
if(head->next==tail){
head->next=nullptr;
return head;
}
ListNode *slow = head;
ListNode *fast = head;
while (fast != tail) {
slow = slow->next;
fast = fast->next;
if (fast != tail) {
fast = fast->next;
}
}
return mergeTwoLists(sortList(head, slow), sortList(slow, tail));
}
ListNode *mergeTwoLists(ListNode *list1, ListNode *list2) {
auto dummy = make_unique<ListNode>(-1);
ListNode *node = dummy.get();
while (list1 != nullptr && list2 != nullptr) {
if (list1->val <= list2->val) {
node->next = list1;
list1 = list1->next;
node = node->next;
} else {
node->next = list2;
list2 = list2->next;
node = node->next;
}
}
while (list1 != nullptr) {
node->next = list1;
list1 = list1->next;
node = node->next;
}
while (list2 != nullptr) {
node->next = list2;
list2 = list2->next;
node = node->next;
}
return dummy->next;
}
};
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