题目:
Sort a linked list in O(n log n) time using constant space complexity.
Example
Given 1->3->2->null
, sort it to 1->2->3->null
.
题解:
O(n log n) : 快速排序,归并排序,堆排序
Solution 1 ()
class Solution {
public:
ListNode *sortList(ListNode *head) {
if (!head || !head->next) {
return head;
} ListNode* slow = head;
ListNode* fast = head->next; while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
}
ListNode* right = sortList(slow->next);
slow->next = nullptr;
ListNode* left = sortList(head); return merge(left, right);
} ListNode* merge(ListNode* l1, ListNode* l2) {
ListNode* dummy = new ListNode(-);
ListNode* cur = dummy;
while (l1 && l2) {
if (l1->val < l2->val) {
cur->next = l1;
l1 = l1->next;
} else {
cur->next = l2;
l2 = l2->next;
}
cur = cur->next;
} if (l1) cur->next = l1;
if (l2) cur->next = l2;
return dummy->next;
}
};
Solutin 2 ()
class Solution {
public:
ListNode* sortList(ListNode* head) {
if (!head || !head->next) return head;
ListNode *slow = head, *fast = head, *pre = head;
while (fast && fast->next) {
pre = slow;
slow = slow->next;
fast = fast->next->next;
}
pre->next = NULL;
return merge(sortList(head), sortList(slow));
}
ListNode* merge(ListNode* l1, ListNode* l2) {
if (!l1) return l2;
if (!l2) return l1;
if (l1->val < l2->val) {
l1->next = merge(l1->next, l2);
return l1;
} else {
l2->next = merge(l1, l2->next);
return l2;
}
}
};