Massive Algorithms: Sort LinkedList

Sort LinkedList

http://yuanhsh.iteye.com/blog/2186111
http://www.cnblogs.com/TenosDoIt/p/3666585.html
归并排序（算法交换链表节点，时间复杂度O（nlogn）,不考虑递归栈空间的话空间复杂度是O（1））
首先用快慢指针的方法找到链表中间节点，然后递归的对两个子链表排序，把两个排好序的子链表合并成一条有序的链表。归并排序应该算是链表排序最佳的选择了，保证了最好和最坏时间复杂度都是nlogn，而且它在数组排序中广受诟病的空间复杂度在链表排序中也从O(n)降到了O(1)

    ListNode *mergeSortList(ListNode *head) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

        //链表归并排序

        if(head == NULL || head->next == NULL)return head;

        else

{

            //快慢指针找到中间节点

            ListNode *fast = head,*slow = head;

            while(fast->next != NULL && fast->next->next != NULL)

{

                fast = fast->next->next;

                slow = slow->next;

}

            fast = slow;

            slow = slow->next;

            fast->next = NULL;

            fast = sortList(head);//前半段排序

            slow = sortList(slow);//后半段排序

            return merge(fast,slow);

}

}

    // merge two sorted list to one

    ListNode *merge(ListNode *head1, ListNode *head2)

{

        if(head1 == NULL)return head2;

        if(head2 == NULL)return head1;

        ListNode *res , *p ;

        if(head1->val < head2->val)

            {res = head1; head1 = head1->next;}

        else{res = head2; head2 = head2->next;}

        p = res;

        while(head1 != NULL && head2 != NULL)

{

            if(head1->val < head2->val)

{

                p->next = head1;

                head1 = head1->next;

}

            else

{

                p->next = head2;

                head2 = head2->next;

}

            p = p->next;

}

        if(head1 != NULL)p->next = head1;

        else if(head2 != NULL)p->next = head2;

        return res;

}

http://www.dontforgettothink.com/2011/11/23/merge-sort-of-linked-list/

    public Node merge_sort(Node head) {
        if(head == null || head.next == null) { return head; }
        Node middle = getMiddle(head);      //get the middle of the list
        Node sHalf = middle.next; middle.next = null;   //split the list into two halfs

        return merge(merge_sort(head),merge_sort(sHalf));  /recurse on that 
    }

    public Node merge(Node a, Node b) {
        Node dummyHead, curr; dummyHead = new Node(); curr = dummyHead;
        while(a !=null && b!= null) {
            if(a.info <= b.info) { curr.next = a; a = a.next; }
            else { curr.next = b; b = b.next; }
            curr = curr.next;
        }
        curr.next = (a == null) ? b : a;
        return dummyHead.next;
    }

    public Node getMiddle(Node head) {
        if(head == null) { return head; }
        Node slow, fast; slow = fast = head;
        while(fast.next != null && fast.next.next != null) {
            slow = slow.next; fast = fast.next.next;
        }
        return slow;
    }

http://www.dontforgettothink.com/2011/11/23/merge-sort-of-linked-list/
http://bookshadow.com/weblog/2014/11/21/java-merge-sort-of-linked-list/
public Node merge_sort(Node head) { if (head == null || head.next == null) { return head; } Node middle = getMiddle(head); // get the middle of the list Node sHalf = middle.next; middle.next = null; // split the list into two halfs return merge(merge_sort(head), merge_sort(sHalf)); // recurse on that } // Merge subroutine to merge two sorted lists public Node merge(Node a, Node b) { Node dummyHead, curr; dummyHead = new Node(); curr = dummyHead; while (a != null && b != null) { if (a.val <= b.val) { curr.next = a; a = a.next; } else { curr.next = b; b = b.next; } curr = curr.next; } curr.next = (a == null) ? b : a; return dummyHead.next; } // Finding the middle element of the list for splitting public Node getMiddle(Node head) { if (head == null) { return head; } Node slow, fast; //“快慢指针” slow = fast = head; while (fast.next != null && fast.next.next != null) { slow = slow.next; fast = fast.next.next; } return slow; }

快速排序1（算法只交换节点的val值，平均时间复杂度O（nlogn）,不考虑递归栈空间的话空间复杂度是O（1））

这里的partition我们参考数组快排partition的第二种写法(选取第一个元素作为枢纽元的版本，因为链表选择最后一元素需要遍历一遍)，具体可以参考here

这里我们还需要注意的一点是数组的partition两个参数分别代表数组的起始位置，两边都是闭区间，这样在排序的主函数中：

void quicksort(vector<int>&arr, int low, int high)

{

if(low < high)

{

   int middle = mypartition(arr, low, high);

   quicksort(arr, low, middle-1);

   quicksort(arr, middle+1, high);

}

}

对左边子数组排序时，子数组右边界是middle-1，如果链表也按这种两边都是闭区间的话，找到分割后枢纽元middle，找到middle-1还得再次遍历数组，因此链表的partition采用前闭后开的区间（这样排序主函数也需要前闭后开区间），这样就可以避免上述问题

    ListNode *quickSortList(ListNode *head) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

        //链表快速排序

        if(head == NULL || head->next == NULL)return head;

        qsortList(head, NULL);

        return head;

}

    void qsortList(ListNode*head, ListNode*tail)

{

        //链表范围是[low, high)

        if(head != tail && head->next != tail)

{

            ListNode* mid = partitionList(head, tail);

            qsortList(head, mid);

            qsortList(mid->next, tail);

}

}

    ListNode* partitionList(ListNode*low, ListNode*high)

{

        //链表范围是[low, high)

        int key = low->val;

        ListNode* loc = low;

        for(ListNode*i = low->next; i != high; i = i->next)

            if(i->val < key)

{

                loc = loc->next;

                swap(i->val, loc->val);

}

        swap(loc->val, low->val);

        return loc;

}

快速排序2（算法交换链表节点，平均时间复杂度O（nlogn）,不考虑递归栈空间的话空间复杂度是O（1））

这里的partition，我们选取第一个节点作为枢纽元，然后把小于枢纽的节点放到一个链中，把不小于枢纽的及节点放到另一个链中，最后把两条链以及枢纽连接成一条链。

这里我们需要注意的是，1.在对一条子链进行partition时，由于节点的顺序都打乱了，所以得保正重新组合成一条新链表时，要和该子链表的前后部分连接起来，因此我们的partition传入三个参数，除了子链表的范围（也是前闭后开区间），还要传入子链表头结点的前驱；2.partition后链表的头结点可能已经改变

    ListNode *quickSortList(ListNode *head) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

        //链表快速排序

        if(head == NULL || head->next == NULL)return head;

        ListNode tmpHead(0); tmpHead.next = head;

        qsortList(&tmpHead, head, NULL);

        return tmpHead.next;

}

    void qsortList(ListNode *headPre, ListNode*head, ListNode*tail)

{

        //链表范围是[low, high)

        if(head != tail && head->next != tail)

{

            ListNode* mid = partitionList(headPre, head, tail);//注意这里head可能不再指向链表头了

            qsortList(headPre, headPre->next, mid);

            qsortList(mid, mid->next, tail);

}

}

    ListNode* partitionList(ListNode* lowPre, ListNode* low, ListNode* high)

{

        //链表范围是[low, high)

        int key = low->val;

        ListNode node1(0), node2(0);//比key小的链的头结点，比key大的链的头结点

        ListNode* little = &node1, *big = &node2;

        for(ListNode*i = low->next; i != high; i = i->next)

            if(i->val < key)

{

                little->next = i;

                little = i;

}

            else

{

                big->next = i;

                big = i;

}

        big->next = high;//保证子链表[low,high)和后面的部分连接

        little->next = low;

        low->next = node2.next;

        lowPre->next = node1.next;//为了保证子链表[low,high)和前面的部分连接

        return low;

}

插入排序（算法中是直接交换节点，时间复杂度O（n^2）,空间复杂度O（1））

    ListNode *insertionSortList(ListNode *head) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

        if(head == NULL || head->next == NULL)return head;

        ListNode *p = head->next, *pstart = new ListNode(0), *pend = head;

        pstart->next = head; //为了操作方便，添加一个头结点

        while(p != NULL)

{

            ListNode *tmp = pstart->next, *pre = pstart;

            while(tmp != p && p->val >= tmp->val) //找到插入位置

                {tmp = tmp->next; pre = pre->next;}

            if(tmp == p)pend = p;

            else

{

                pend->next = p->next;

                p->next = tmp;

                pre->next = p;

}

            p = pend->next;

}

        head = pstart->next;

        delete pstart;

        return head;

}

http://geeksquiz.com/insertion-sort-for-singly-linked-list/

void insertionSort(struct node **head_ref)

{

    // Initialize sorted linked list

    struct node *sorted = NULL;

    // Traverse the given linked list and insert every

    // node to sorted

    struct node *current = *head_ref;

    while (current != NULL)

{

        // Store next for next iteration

        struct node *next = current->next;

        // insert current in sorted linked list

        sortedInsert(&sorted, current);

        // Update current

        current = next;

}

    // Update head_ref to point to sorted linked list

    *head_ref = sorted;

}

/* function to insert a new_node in a list. Note that this

  function expects a pointer to head_ref as this can modify the

  head of the input linked list (similar to push())*/

void sortedInsert(struct node** head_ref, struct node* new_node)

{

    struct node* current;

    /* Special case for the head end */

    if (*head_ref == NULL || (*head_ref)->data >= new_node->data)

{

        new_node->next = *head_ref;

        *head_ref = new_node;

}

    else

{

        /* Locate the node before the point of insertion */

        current = *head_ref;

        while (current->next!=NULL &&

               current->next->data < new_node->data)

{

            current = current->next;

}

        new_node->next = current->next;

        current->next = new_node;

}

}

选择排序（算法中只是交换节点的val值，时间复杂度O（n^2）,空间复杂度O（1））

    ListNode *selectSortList(ListNode *head) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

        //选择排序

        if(head == NULL || head->next == NULL)return head;

        ListNode *pstart = new ListNode(0);

        pstart->next = head; //为了操作方便，添加一个头结点

        ListNode*sortedTail = pstart;//指向已排好序的部分的尾部

        while(sortedTail->next != NULL)

{

            ListNode*minNode = sortedTail->next, *p = sortedTail->next->next;

            //寻找未排序部分的最小节点

            while(p != NULL)

{

                if(p->val < minNode->val)

                    minNode = p;

                p = p->next;

}

            swap(minNode->val, sortedTail->next->val);

            sortedTail = sortedTail->next;

}

        head = pstart->next;

        delete pstart;

        return head;

}

冒泡排序（算法交换链表节点val值，时间复杂度O（n^2）,空间复杂度O（1））

    ListNode *bubbleSortList(ListNode *head) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

        //链表快速排序

        if(head == NULL || head->next == NULL)return head;

        ListNode *p = NULL;

        bool isChange = true;

        while(p != head->next && isChange)

{

            ListNode *q = head;

            isChange = false;//标志当前这一轮中又没有发生元素交换，如果没有则表示数组已经有序

            for(; q->next && q->next != p; q = q->next)

{

                if(q->val > q->next->val)

{

                    swap(q->val, q->next->val);

                    isChange = true;

}

}

            p = q;

}

        return head;

}

对于希尔排序，因为排序过程中经常涉及到arr[i+gap]操作，其中gap为希尔排序的当前步长，这种操作不适合链表。

对于堆排序，一般是用数组来实现二叉堆，当然可以用二叉树来实现，但是这么做太麻烦，还得花费额外的空间构建二叉树

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