LeetCode 147. Insertion Sort List

Solution1：我的答案
有点笨，有点慢

/*** 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) return NULL;if (!head->next) return head;ListNode* new_head = new ListNode(-1), *cur = head;while (cur) {ListNode *temp = cur->next;cur->next = NULL;my_insert(new_head, cur);cur = temp;}return new_head->next;}void my_insert (ListNode* &new_head, ListNode* &des) {if (!new_head->next || des->val <= new_head->next->val) {ListNode *temp = new_head->next;new_head->next = des;des->next = temp;return;} else {ListNode* cur = new_head->next;while (cur->next) {if (des->val >= cur->val && des->val <= cur->next->val) {des->next = cur->next;cur->next = des;return;}elsecur = cur->next;}cur->next = des;return;}}
};

Solution2：
参考网址：http://www.cnblogs.com/grandyang/p/4250107.html

/*** 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) {ListNode *dummy = new ListNode(-1), *cur = dummy;while (head) {ListNode *t = head->next;//暂时保存头结点的下一个位置cur = dummy;while (cur->next && cur->next->val <= head->val) {cur = cur->next;}head->next = cur->next;cur->next = head;head = t;}return dummy->next;}
};

要反思为啥别人的代码写的如此简洁~~~

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