简单的链表排序的问题

题目：如，给出 1->3->2->null，给它排序变成 1->2->3->null.

写了简单的答案不能通过求帮忙看下错误原因

第一次提问这个格式怎么编辑o(╯□╰)o

/**

• Definition of ListNode
• class ListNode {
• public:
• int val;
• ListNode *next;
• ListNode(int val) {
• this->val = val;
• this->next = NULL;
• }
• } / class Solution { public: /*
  • @param head: The first node of linked list.
  • @return: You should return the head of the sorted linked list, using constant space complexity. */ ListNode *sortList(ListNode *head) { // write your code here if(head==NULL) return NULL; if(head->next==NULL) return head; ListNode *dummy = new ListNode(0); ListNode *temp = dummy; vector nums; while(head!=NULL) { nums.push_back(head->val); head = head->next; } sort(nums.begin(),nums.end()); for(int i = 0;i < nums.size();i++) { temp->next->val = nums[i]; temp = temp->next; } temp->next = NULL; return dummy->next; } };

http://blog.csdn.net/a29562268/article/details/53932682
参考这个insear方法，把要动的节点前后保存下，然后再添加就行