[leetcode]Insertion Sort List

给一个链表,用插入排序...

就是考链表的基本操作...

开始想法很2,重新new一个链表...

但是我明明都有链表了,我去new干嘛呢...

其实嘛，排序就是把链表重新连下就好啦，一个一个的独立块，把next重新连下

要是不做这题我还真不会这么去做，平时链表用的太少了T_T

还有就是单向链表，要在当前位置插入东西的话要记录他的上一位置才行哟。。。

 

/**
 * 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 == nullptr) return head;
        
        ListNode* curr = head -> next;
        head -> next = nullptr;
        
        while(curr != nullptr){
            ListNode* tmpHead = head;
            ListNode* prev = nullptr;
            ListNode* next = curr -> next;
            while(tmpHead != nullptr && tmpHead -> val <= curr -> val)
            {
                prev = tmpHead;
                tmpHead = tmpHead -> next;
            }
            
            if(prev != nullptr){
                //insert
                if(prev -> next){
                    curr -> next = prev -> next;
                    prev -> next = curr;
                }else
                {
                    prev -> next = curr;
                    curr -> next = nullptr;
                }
            }else{
                curr -> next = head;
                head = curr;
            }
            curr = next;
        }
        return head;
    }
    
};

 

/**
 * 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 == nullptr) return head;
        ListNode* sort_head = head;
        ListNode* prev = nullptr;
        head = head->next;
        sort_head->next = nullptr;
        while(head) {
            prev = nullptr;
            ListNode* tmp = sort_head;
            ListNode* now = head;
            ListNode* next = head->next;
            while(tmp) {
                if (tmp->val > now->val) break;
                prev = tmp;
                tmp = tmp->next;
            }
            if (prev) {
                ListNode* next = prev->next;
                prev->next = now;
                now->next = next;
            } else {
                now->next = sort_head;
                sort_head = now;
            }
            head = next;
        }
        return sort_head;
    }
};