【LeetCode】19. Remove Nth Node From End of List

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.

题意：给出一个链表，删除倒数第n个节点

刚在书上见过这个题，思路是用双指针，p1,p2，先让p2走n步，然后p1,p2一块走，当p2停止的时候，

p1->next正好指向要删除的节点，不过要注意：

当p2停止的时候，n还没变为0，这时候要验证n的值，当n>0时候，p1指向的节点就是要删除的节点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    if(head==NULL)
        return head;
    struct ListNode *p1,*p2,*tmp;
    p1=head;
    p2=head;
    while(n>0&&p2->next!=NULL){
        p2=p2->next;
        n--;
    }
    if(n>0){
        tmp=p1;
        head=p1->next;
        free(tmp);
    }
    else
    {
    while(p2->next!=NULL){
        p2=p2->next;
        p1=p1->next;
    }
    tmp=p1->next;
    if(tmp!=NULL){
        p1->next=tmp->next;
        free(tmp);
    }
    }
    return head;
}