LeetCode：Sort List

Title：

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

思路：考虑快速排序和归并排序，但是我的快速排序超时了

ListNode* sortList(ListNode* head) {
        if (!head)
            return NULL;
        int m_val = head->val;
        ListNode* left = NULL,*right = NULL;
        ListNode* p_left = NULL,*p_right = NULL;
        ListNode* p = head->next;
        if (p == NULL)
            return head;
        while (p){
            if (p->val < m_val){
                if (left == NULL)
                    left = p;
                else
                    p_left->next = p;
                p_left = p;
            }else{
                if (right == NULL)
                    right = p;
                else
                    p_right->next = p;
                p_right = p;
            }
            p = p->next;
        }
        if (p_left)
            p_left->next = NULL;
        if (p_right)
            p_right->next = NULL;
        ListNode* sorted_left = sortList(left);
        ListNode* sorted_right = sortList(right);
        ListNode *sorted = new ListNode(0);
        p = sorted_left;
        ListNode* p_cur = sorted;
        while (p){
            p_cur->next = p;
            p_cur = p;
            p = p->next;
        }
        p_cur->next = head;
        p_cur = head;
        p_cur->next = sorted_right;
        return sorted->next;
    }

归并排序。两种思路，一个得到全部长度之后，进行。另一种使用快慢指针

class Solution {
private:
    int len(ListNode* head){
        int l = 0;
        ListNode*p = head;
        while (p){
            l++;
            p = p->next;
        }
        return l;
    }

public:
    ListNode* sortList(ListNode* head){
        if (head == NULL)
            return NULL;
        int length = len(head);
        
        return sort(head,length);
    }
    ListNode* sort(ListNode*& head,int length){
        if (length == 1){
            ListNode* t = head;
            head = head->next;//这个地方要注意。同时使用指针引用的目的是让head一直往后。因为总是左边的先走，所以到sort(head,length-length/2)时head刚好就是那个位置。
            t->next =  NULL;
            return t;
        }
        ListNode*left = sort(head,length/2);
        ListNode*right = sort(head,length-length/2);
        return merge(left,right);
    }
    ListNode* merge(ListNode* left,ListNode* right){
        ListNode* head = new ListNode(0);
        ListNode* p = head;
        while (left && right){
            if (left->val < right->val){
                p->next = left;
                p = left;
                left = left->next;
            }else{
                p->next = right;
                p = right;
                right = right->next;
            }
        }
        while (left){
            p->next = left;
            p = left;
            left = left->next;
        }
        while (right){
            p->next = right;
            p = right;
            right = right->next;
        }
        return head->next;
        
    }
}

class Solution {
public:
    ListNode *sortList(ListNode *head) {
        if(!head||!head->next)
            return head;
        return mergeSort(head);
    }
    ListNode * mergeSort(ListNode *head){
        if(!head||!head->next)   //just one element
            return head;
        ListNode *p=head, *q=head, *pre=NULL;
        while(q&&q->next!=NULL){
            q=q->next->next;
            pre=p;
            p=p->next;  //divide into two parts
        }
        pre->next=NULL;
        ListNode *lhalf=mergeSort(head);
        ListNode *rhalf=mergeSort(p);  //recursive
        return merge(lhalf, rhalf);   //merge
    }
    ListNode * merge(ListNode *lh, ListNode *rh){
        ListNode *temp=new ListNode(0);
        ListNode *p=temp;
        while(lh&&rh){
            if(lh->val<=rh->val){
                p->next=lh;
                lh=lh->next;
            }
            else{
                p->next=rh;
                rh=rh->next;
            }
            p=p->next;
        }
        if(!lh)
            p->next=rh;
        else
            p->next=lh;
        p=temp->next;
        temp->next=NULL;
        delete temp;
        return p;
    }
};