leetcode:Remove Nth Node From End of List
2015-09-21 20:32
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题目
给定一个列表,删除倒数第n个节点,并且返回表头,例如:给定列表:1->2->3->4->5, n = 2
返回:1->2->3->5
假设给定的n值都是合法的。
注意
给定空表,返回NULL如果删除第一个节点,要保证不要删了表头
思路
设置两个指针p,q,相差n−1,那么如果后一个q指针知道表尾时,前一个指针p所在位置即为要删除的节点,记录p的直接前驱,删除p即可。struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: ListNode* removeNthFromEnd(ListNode* head, int n) { if (NULL == head) {return head;} ListNode *p = head; ListNode *q = head; ListNode *pPre = NULL;//记录 p 指针直接前驱 for(int i = 0;i != n - 1;++i) q = q->next; while(q->next) { pPre = p; q = q->next; p = p->next; } if(NULL == pPre)//要删除的节点是第一个节点 { head = p->next; delete p; } else { pPre->next = p->next; delete p; } return head; } };
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