[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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