Leetcode: 19. Remove Nth Node From End of List

Description

Given a linked list, remove the nth node from the end of list and return its head.

Example

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.

思路

• 边界情况主要最第一个和最后一个的处理
• 先遍历链表，找出链表长度
• 然后计算出要删除的节点的前一个节点
• 此时，需要分清楚是否是删除第一个节点。然后就没有然后了

代码

/**
 * 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) return NULL;
        
        ListNode *ptr = head;
        int len = 0;
        
        while(ptr){
            ptr = ptr->next;
            len++;
        }

        ptr = head;
        int tmp = len - n;
        n = tmp;
        cout << tmp << endl;
        while(tmp > 1){
            ptr = ptr->next;
            tmp--;
        }      
        
        if(n > 0){
            if(ptr->next){
                ListNode *tmp = ptr->next;
                ptr->next = tmp->next;
                delete tmp;
            }
            else ptr->next = NULL;
        }
        else
        {
            ListNode *tmp = head;
            head = head->next;
            delete tmp;
        }
        
        return head;
    }
};