LeetCode 19. Remove Nth Node From End of List

题目描述:

Given a linked list, remove the n^th 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.

要求删除链表从后往前数第n个节点, 并且只遍历一次. 所以我采用一个vector来保存每个节点的指针, 遍历一次后从vector中找到倒数第n个节点并删除之.

/**
* 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) {
        vector<ListNode*> listPointer;
        ListNode* p = head;
        while(p != nullptr){
            listPointer.push_back(p);
            p = p->next;
        }
        
        int len = listPointer.size(), toDeleteIndex = len - n;
        ListNode* toDelete = listPointer[toDeleteIndex];
        if(toDeleteIndex == 0){
            return head->next;
        }
        else{
            ListNode* prev = listPointer[toDeleteIndex - 1];
            prev->next = toDelete->next;
            return head;
        }
    }
};