Sort List
描述
Sort a linked list in O(n log n)
time using constant space complexity.
分析
常数空间且O(nlogn)
,单链表适合用归并排序,双向链表适合用快速排序。本题可以复用 Merge Two Sorted Lists 的代码。
代码
// Sort List
// 归并排序,时间复杂度O(nlogn),空间复杂度O(1)
class Solution {
public:
ListNode *sortList(ListNode *head) {
if (head == NULL || head->next == NULL)return head;
ListNode *middle = findMiddle(head);
ListNode *head2 = middle->next;
middle->next = nullptr; // 断开
ListNode *l1 = sortList(head); // 前半段排序
ListNode *l2 = sortList(head2); // 后半段排序
return mergeTwoLists(l1, l2);
}
private:
// Merge Two Sorted Lists
static ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
ListNode dummy(-1);
for (ListNode* p = &dummy; l1 != nullptr || l2 != nullptr; p = p->next) {
int val1 = l1 == nullptr ? INT_MAX : l1->val;
int val2 = l2 == nullptr ? INT_MAX : l2->val;
if (val1 <= val2) {
p->next = l1;
l1 = l1->next;
} else {
p->next = l2;
l2 = l2->next;
}
}
return dummy.next;
}
// 快慢指针找到中间节点
static ListNode* findMiddle(ListNode* head) {
if (head == nullptr) return nullptr;
ListNode *slow = head;
ListNode *fast = head->next;
while (fast != nullptr && fast->next != nullptr) {
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
};