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148. Sort List
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/** | |
* Definition for singly-linked list. | |
* struct ListNode { | |
* int val; | |
* ListNode *next; | |
* ListNode(int x) : val(x), next(NULL) {} | |
* }; | |
*/ | |
class Solution { | |
public: | |
ListNode* sortList(ListNode* head) { | |
if(head == NULL || head->next == NULL){ | |
return head; | |
} | |
// 使用 merge sort 的 top down 方法,才符合提題目 | |
// time : O(nlogn) | |
// space : O(1) | |
// top down 是將 list 拆分成個位數,然後在兩兩合併 | |
// 1. 將 list 拆分成 left(l1) 與 right(l2) | |
ListNode* slow = head; | |
ListNode* fast = head->next; | |
ListNode* l1; | |
ListNode* l2; | |
while(fast && fast->next){ | |
slow = slow->next; | |
fast = fast->next->next; | |
} | |
ListNode* temp = slow->next; | |
slow->next = NULL; | |
l1 = head; | |
l2 = temp; | |
// merge l1 與 l2 | |
return merge(sortList(l1), sortList(l2)); | |
} | |
private: | |
ListNode* merge(ListNode* l1, ListNode* l2){ | |
// 會需要一個 temp 的原因,如果沒有它 dummy 會需要一直改變 | |
// ex. dummy = dummy->next; | |
// 所以需要一個 temp 來替代它。 | |
ListNode* dummy = new ListNode(-1); | |
ListNode* temp = dummy; | |
while(l1 && l2){ | |
if(l1->val < l2->val){ | |
temp->next = l1; | |
l1 = l1->next; | |
}else{ | |
temp->next = l2; | |
l2 = l2->next; | |
} | |
temp = temp->next; | |
} | |
if(l1) temp->next = l1; | |
if(l2) temp->next = l2; | |
return dummy->next; | |
} | |
}; |
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