【Remove Nth Node From End of List】cpp

题目：

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.

代码：9ms过集合

/**
 * 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 head;
        ListNode dummy(-1);
        dummy.next = head;
        ListNode *p1 = &dummy, *p2 = &dummy;
        for (size_t i = 0; i < n; ++i, p2=p2->next);
        for (; p2->next; p1=p1->next, p2=p2->next);
        ListNode *tmp = p1->next;
        p1->next = p1->next==NULL ? NULL : p1->next->next;
        delete tmp;
        return dummy.next;
    }
};

Tips:

双指针技巧：

　　a. p2先走n步

　　b. p1和p2一起走，直到p2走到最后一个元素

　　c. 删除元素

注意再删除元素的时候，保护一下p1->next指针不为NULL。

==========================================

第二次过这道题，思路比较清晰。由于受到Rotate List这道题的影响，第一次把p1 = head 和 p2 = head了；之后改成了p1 = &dummpy和p2 = &dummpy就AC了。

/**
 * 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 head;
            ListNode dummpy(-1);
            dummpy.next = head;
            ListNode* p1 = &dummpy;
            ListNode* p2 = &dummpy;
            for ( int i=0; i<n; ++i ) p2 = p2->next;
            while ( p2 && p2->next)
            {
                p1 = p1->next;
                p2 = p2->next;
            }
            p1->next = p1->next ? p1->next->next : NULL;
            return dummpy.next;
    }
};