leetcode--Remove Nth Node From End of List

1.题目分析

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.

2.解法分析

对于链表的题,双指针法经常用到,这个题目就可以设置两个指针,题意说n总是有效,可以省去很多判断,但是为了解题严谨,我还是加上了很多防止异常的检测.

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(n<=0)return head;
        if(head==NULL)return NULL;

        ListNode *forward=head;


        int margin=0;
        while(margin<n&&forward!=NULL)
        {
            margin++;
            forward=forward->next;
        }

        if(margin<n)return head;

        ListNode *afterward=head;
        ListNode *prev=head;
        while(forward!=NULL)
        {
            prev=afterward;
            afterward=afterward->next;
            forward=forward->next;
        }

        if(afterward==head)
        {
            prev=head;
            afterward=head->next;
            free(head);
            return afterward;
        }

        prev->next=afterward->next;
        free(afterward);
        return head;


    }
};