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

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.

Follow up:

Could you do this in one pass? 你能一遍过吗（笑

AC代码

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head->next == NULL) return NULL;
        ListNode *head_cp = head, *shadow = head, *pre_shadow = head;
        n--;
        while (n--) head_cp = head_cp->next;
        while (head_cp->next != NULL) {
            head_cp = head_cp->next;
            pre_shadow = shadow;
            shadow = shadow->next;
        }
        if (pre_shadow != shadow) {
            pre_shadow->next = shadow->next;
            delete shadow;
        }
        else {
            ListNode* t = shadow->next;
            delete shadow;
            return t;
        }
        return head;
    }
};

总结

无