147. Insertion Sort List

Sort a linked list using insertion sort.

解法：

Well, life gets difficult pretty soon whenever the same operation on array is transferred to linked list.

First, a quick recap of insertion sort:

Start from the second element (simply a[1] in array and the annoying head -> next -> val in linked list), each time when we see a node with val smaller than its previous node, we scan from the head and find the position that the current node should be inserted. Since a node may be inserted before head, we create a new_head that points to head. The insertion operation, however, is a little easier for linked list.

Now comes the code:

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* insertionSortList(ListNode* head) {        if(!head||!head->next) return head;        ListNode* new_head ;        new_head->next = head;        ListNode* pre = new_head;        ListNode* cur = head;        while(cur) {            if(cur->next && cur->next->val < cur->val) {                while(pre->next && pre->next->val < cur->next->val) pre = pre->next;                ListNode* temp = pre->next;                pre->next = cur->next;                cur->next = cur->next->next;                pre->next->next = temp;                pre = new_head;            }            else cur = cur->next;        }        return new_head->next;    }};

方法二：

class Solution {public:    /**     * @param head: The first node of linked list.     * @return: The head of linked list.     */    ListNode *insertionSortList(ListNode *head) {        ListNode *dummy = new ListNode(0);        // 这个dummy的作用是，把head开头的链表一个个的插入到dummy开头的链表里        // 所以这里不需要dummy->next = head;        while (head != NULL) {            ListNode *temp = dummy;            ListNode *next = head->next;            while (temp->next != NULL && temp->next->val < head->val) {                temp = temp->next;            }            head->next = temp->next;            temp->next = head;            head = next;        }        return dummy->next;    }};