這個題目比較有意思,因爲平時比較熟悉的排序一般就是對數組操作,題目要求O(n log n)和constant的空間複雜度,能有O(n log n)的算法一般就是merge sort,quick sort或者heap sort,這裏quick sort難以實現因爲核心步驟partion需要雙向的指針,所以quick sort適合雙向鏈表,這裏我用merge sort,一個比較重要的方法就是用O(n)的複雜度找到鏈表中心節點,用的快慢指針即可實現,然後就這樣不停的分解鏈表然後merge,代碼如下:
/**
* 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;
ListNode* p = head;
ListNode* q = head;
ListNode* pre = NULL;
while(q&&q->next){
pre = p;
p = p->next;
q = q->next->next;
}
pre->next = NULL;
ListNode* left = sortList(head);
ListNode* right = sortList(p);
return mergeList(left,right);
}
private:
static ListNode* mergeList(ListNode* left,ListNode* right){
ListNode* head = new ListNode(-1);
ListNode *p = head;
while(left&&right){
if(left->val>right->val){
p->next = right;
right = right->next;
}
else{
p->next = left;
left = left->next;
}
p = p->next;
}
if(left){
p->next = left;
}
if(right){
p->next = right;
}
return head->next;
}
};