Leetcode: Remove Nth Node From End of List 题目: Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->...
用慢指针指向要移除的Node的前一个Node. 2. 使用dummy node作为head的前缀节点,这样就算是删除head也能轻松handle啦! 主页君是不是很聪明呀? :) View Code GITHUB (国内用户可能无法连接): https://github.com/yuzhangcmu/LeetCode/blob/251766ffb832f2278f43a05e194ca76584bf14ea/list/RemoveNthFromEnd.ja...
Given1->1->1->2->3, return2->3. 解题思路:防止情况的多样性,先在head前面加上一个节点node,这样无论什么情形都可以统一为一种处理方式。 设置一个work指针指向node,如果work->next->val == work->next->next->val,表明有相同元素出现,则在内循环内将相同元素删除到只剩2个,在循环结束时,再讲这2...
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LeetCode 938. Range Sum of BST 2019-12-12 08:00 −原题链接在这里:https://leetcode.com/problems/range-sum-of-bst/ 题目: Given the root node of a binary search tree, return the sum of values of all node... Dylan_Java_NYC
Recover a Tree From Preorder Traversal 1027. Longest Arithmetic Sequence 1026. Maximum Difference Between Node and Ancestor 1025. Divisor Game 1024. Video Stitching 1023. Camelcase Matching 1022. Sum of Root To Leaf Binary Numbers 1021. Remove Outermost Parentheses 1020. Number of Enclaves 1019. ...
2.https://leetcode.com/problems/remove-zero-sum-consecutive-nodes-from-linked-list/discuss/549912/C%2B%2B-8ms 删除之后,继续遍历时,删除的情况比如:-3 3 -2 2 classSolution {public: ListNode* removeZeroSumSublists(ListNode*head) { ListNode ln(0);ln.next=head; ...
Given1->1->2->3->3, return1->2->3. 顺序扫描一遍即可,注意释放被删除的结点的空间。 /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {}
http://oj.leetcode.com/problems/remove-duplicates-from-sorted-array-ii/ 要求可以用重复元素,但是最大重复次数为两次。 #include <iostream>usingnamespacestd;classSolution {public:intremoveDuplicates(intA[],intn) {if(n ==0)return0;if(n ==1)return1;inttimes =1;intpreOne = A[0];intnextAvali...
[LeetCode]Remove Nth Node From End of List publicclassSolution {publicListNode removeNthFromEnd(ListNode head,intn) { ListNode dummy=newListNode(0); dummy.next=head; head=dummy; ListNode pre=head; ListNode p=pre.next;for(inti = 0; i < n; i++) {...