链表排序 - Sort LinkedList

这篇文章分析一下链表的各种排序方法。

 

以下排序算法的正确性都可以在LeetCode的链表排序这一题检测。本文用到的链表结构如下（排序算法都是传入链表头指针作为参数，返回排序后的头指针）

struct ListNode {

  int val;

  ListNode *next;

  ListNode(int x) : val(x), next(NULL) {}

  };

 

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））

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class Solution { public:     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;     } };

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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））

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class Solution { public:     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;     } };

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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

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class Solution { public:     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

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class Solution { public:     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;     } };

 

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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））                        本文地址

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)

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class Solution { public:     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;     } };

 

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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））

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class Solution { public:     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;     } };

 

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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

 

文章转自：

http://www.cnblogs.com/TenosDoIt/p/3666585.html