2、题目分析 利用插入排序的算法即可。注意操作指针。 3、代码 1ListNode* insertionSortList(ListNode*head) {2if(head == NULL || head->next ==NULL)3returnhead;45ListNode *dummy =newListNode(0);6dummy->next =head;7ListNode *pre =dummy;8ListNode *p =head;9ListNode *pn = head->next;1011wh...
题解: 1/**2* Definition for singly-linked list.3* public class ListNode {4* int val;5* ListNode next;6* ListNode(int x) {7* val = x;8* next = null;9* }10* }11*/12publicclassSolution {13publicListNode insertionSortList(ListNode head) {14//dummy is dummy head node,not head p...
leetcode -- Insertion Sort List -- 重点,需要优化 https://leetcode.com/problems/insertion-sort-list/ 需要想清楚再写code。终止条件是什么,有哪些变量循环。。。注意加上dummy_node 自己写的code 效率低,可以AC. 复习的时候注意要看看如何优化 class Solution(object): def insertionSortList(self, head): ...
https://leetcode.com/problems/insertion-sort-list/ 题目: Sort a linked list using insertion sort. 思路: 头插法。用头结点可以简化插入链表时候的操作,因为要考虑插入链表中间和表头两种情况,插入表头时,head就要更新,还要判断pre指针是否为空 算法: public ListNode insertSortList(ListNode head, ListNode t)...
总结: sort list using insertion sort ** Anyway, Good luck, Richardo! My code: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */publicclassSolution{publicListNodeinsertionSortList(ListNode head){if...
1. Description Insertion Sort List 2. Solution /** * 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){retu...
[LeetCode] Insertion Sort List 简介:Sort a linked list using insertion sort.解题思路 对于得到结点current的插入位置,从头结点开始遍历,直到遍历到值大于等于节点current的结点,然后将从该结点到current的前驱结点的所有结点的值依次和current结点的值交换,从而达到将该节点插入所遍历到的位置的目的。实现代码/***...
classSolution{ public: ListNode*insertionSortList(ListNode*head){ if(head==NULL||head->next==NULL)//only 0,1nodes { returnhead; } ListNode*pNode=head->next,*hNode=head,*tNode=NULL; hNode->next=NULL; while(pNode) { hNode=head; ...
https://leetcode.com/problems/insertion-sort-list/链表的插入排序。实现起来有不少细节须要仔细考虑。class Solution { public: ListNode* insertionSortList(ListNode* head) { if (!head || !head->next) return head; ListNode dummy(0); dummy.next = head; head = head->next; ListNode *pre = ...
* } */ public class Solution { public ListNode insertionSortList(ListNode head) { if(head == null || head.next == null)return hea 剩余60%内容,订阅专栏后可继续查看/也可单篇购买 小白刷Leetcode文章被收录于专栏 那些必刷的leetcode