Leetcode 19 Remove Nth Node From End of List 超简洁代码
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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->5.
Note:
Given n will always be valid.
Try to do this in one pass.
利用类似DFS的递归解决,自认为还是挺简洁的。
既然是Leetcode上的基础题,就要小题大做,我去看了一下discuss,
发现有人用一种快慢指针的方法,即快指针先走n步,然后两者同时走,当快指针走到尾部时,慢指针就是要删除的节点,想法也挺好的。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: int cnt=0; ListNode* removeNthFromEnd(ListNode* head, int n) { if(head==NULL) return NULL; head->next=removeNthFromEnd(head->next,n); if(++cnt==n) return head->next; return head; }};
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