[LeetCode] Remove Nth Node From End of List
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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.
分析:
依然是双指针思想,两个指针相隔n-1,每次两个指针向后一步,当后面一个指针没有后继了,前面一个指针就是要删除的节点。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public: ListNode* removeNthFromEnd(ListNode* head, int n) { if(head==NULL) return NULL; ListNode *p,*q,*p_pre; p_pre=NULL; p=head; q=head; for (int i=0;i<n-1;i++) {q=q->next; } while (q->next!=NULL) { p_pre=p; p=p->next; q=q->next; } if(p_pre==NULL) { head=p->next; delete p;} else {p_pre->next=p->next; delete p;} return head; }};注意:
1、记得考虑删除的是头节点的情况
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