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

/** * 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 == nullptr || head->next == nullptr) return head;                ListNode * pivot = head, * small = nullptr, * large = nullptr, *sp = nullptr, *lp = nullptr;        head = head->next;        while(head) {            if(head->val < pivot->val) {                if (small == nullptr) small = head;                else sp->next = head;                sp = head;            } else {                if (large == nullptr) large = head;                else lp->next = head;                lp = head;            }            head = head->next;        }        if(sp) sp->next = nullptr;        if(lp) lp->next = nullptr;                // sort the sub list.        sp = sortList(small);        lp = sortList(large);                // merge the sub list.        if(sp)  {            head = sp;            // seek to the last node.            while(sp->next)                sp = sp->next;            sp->next = pivot;        } else {            head = pivot;        }                pivot->next = lp;        return head;    }};

if we only change the value. it will improve a some time complexity.

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {private:    ListNode * partition(ListNode * head, ListNode * tail) {        if(head == tail || head->next == nullptr) return head;        int pivot = head->val;        ListNode * slow = head, *fast = head->next;        while(fast && fast != tail) {            if (fast->val < pivot) {                slow = slow->next;                swap(slow->val, fast->val);            }            fast = fast->next;        }        // swap the pivot        swap(slow->val, head->val);        return slow;    }        void quicksort(ListNode * head, ListNode * tail) {        if(head == tail || head->next == nullptr) return;                ListNode * pivot = partition(head, tail);        quicksort(head, pivot);        quicksort(pivot->next, tail);    }public:    ListNode* sortList(ListNode* head) {        quicksort(head, nullptr);        return head;    }};

if we do a non-recursive merge sort. the effeciency will be time O(NlogN), and the space with O(1).

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {private:    // from head->@->@->@->null to    // head->@->@->null  current->@->null    ListNode * cut(ListNode * head, int n) {        if (head == nullptr) return head;        ListNode * p = head;        while(--n && p) {            p = p->next;        }        if (p == nullptr) return nullptr;        ListNode * next = p->next;        // cut the current node        p->next = nullptr;        return next;    }        // merge two linklist and return the header    ListNode * merge(ListNode * l, ListNode * r) {        ListNode dummyHead(0);        ListNode * p = &dummyHead;                while(l && r) {            if(l->val < r->val) {                p->next = l;                p = l;                l = l->next;            } else {                p->next = r;                p = r;                r = r->next;            }        }        p->next = l ? l : r;        return dummyHead.next;    }    public:    ListNode* sortList(ListNode* head) {        ListNode dummyHead(0);        dummyHead.next = head;                int length = 0;        while(head) {            head = head->next;            ++length;        }                for(int size = 1; size < length; size <<= 1) {            ListNode * cur = dummyHead.next;            ListNode * p = &dummyHead;                        while(cur) {                ListNode * left = cur;                ListNode * right = cut(cur, size);                cur = cut(right, size);                                p->next = merge(left, right);                // seek to the end.                while(p->next) {                    p = p->next;                }                            }        }                return dummyHead.next;            }};

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