题解 | #删除链表的倒数第n个节点#
删除链表的倒数第n个节点
http://www.nowcoder.com/practice/f95dcdafbde44b22a6d741baf71653f6
/**
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
class Solution {
public:
/**
*
* @param head ListNode类
* @param n int整型
* @return ListNode类
*/
ListNode* removeNthFromEnd(ListNode* head, int n) {
// write code here
// 设置双指针,slow和fast
// slow指向head结点前一个节点,fast指向head节点
// 先让fast向前移动n个节点,然后一同移动fast和slow节点
// 当fast节点到达链表末尾时将slow节点前的节点给删去
ListNode* node = new ListNode(-1);
node->next = head;
ListNode* fast = head;
ListNode* slow = node;
while(n-- && fast)
fast = fast->next;
while(fast && slow->next) {
fast = fast->next;
slow = slow->next;
}
slow->next = slow->next->next;
return node->next;
}
};
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