Lintcode098.Sort List
Posted Vincent丶
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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(-1); 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; } } };
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