Sort List

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------QUESTION------

Sort a linked list in O(n log n)time using constant space complexity.

------SOLUTION------

class Solution {public:    ListNode *sortList(ListNode *head) {        if(!head || !head->next ) return head;        return quickSort(head)[0];    }        vector< ListNode* > quickSort(ListNode *head){        vector< ListNode* > ret;        if(!head->next) {            ret.push_back(head);            ret.push_back(head);            return ret;        }        ListNode* headSmaller = NULL;        ListNode* headGreater = NULL;         ListNode* current = head->next;        ListNode* prev = head;        ListNode* tmp;        while(current){            if(current->val > head->val){                prev->next = current->next;                if(!headGreater){                    headGreater = current;                    headGreater->next = NULL;                }                else{                    current->next  = headGreater->next;                    headGreater->next = current;                }                current = prev->next;            }            else if(current->val < head->val){                prev->next = current->next;                if(!headSmaller){                    headSmaller = current;                    headSmaller->next = NULL;                }                else{                    current->next  = headSmaller->next;                    headSmaller->next = current;                }                current = prev->next;            }            else {                prev = current;                current = current->next;            }        }                vector< ListNode* > retGreater;        vector< ListNode* > retSmaller;        if(headSmaller) {            retSmaller = quickSort(headSmaller);            retSmaller[1]->next = head;            ret.push_back(retSmaller[0]);        }        else ret.push_back(head);        if(headGreater){           retGreater = quickSort(headGreater);           prev->next = retGreater[0];           ret.push_back(retGreater[1]);        }        else ret.push_back(prev);        return ret;    }};

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