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)
http://www.dontforgettothink.com/2011/11/23/merge-sort-of-linked-list/
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; } 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; }
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））
http://geeksquiz.com/insertion-sort-for-singly-linked-list/
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））
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））
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：
voidquicksort(vector<int>&arr,intlow,inthigh){if(low < high){intmiddle = 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; }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; }} 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; } 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
