Question: Remove Nth Node From End of List

Description: 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: n will always be valid.

Solution Code:

class Solution {
    /**
     * 注意以下幾點：
     * (1)提示： n will always be valid. 所以不需要考慮n<=0 或者超過連結串列長度的問題
     * (2)若待刪除節點不是頭結點，則不需要維護頭結點
     * (3)待刪除節點是頭結點的情況，需要注意
     */
public:
    ListNode* removeNthFromEnd(ListNode* head, int
 
 n) {
        if (head == NULL)
            return NULL;

        ListNode* pRight = head;
        ListNode* pLeft = head;
        ListNode* pHead = head;
        int count = 0;
        while(pRight != NULL && count < n){
            pRight = pRight->next;
            ++count;
        }
        
 
// 待刪除節點是頭結點的情況
        if(pRight == NULL){
            pHead = head->next;
            free(head);
            return pHead;
        }
        // 查詢待刪除節點的前一個節點
        while(pRight->next != NULL){
            pRight = pRight->next;
            pLeft = pLeft->next;
        }
        // 執行刪除
        ListNode* pDelete = pLeft->next;
        pLeft->next = pLeft->next->next;
        free(pDelete);
        return pHead;
    }
};

Reports: Runtime: 4 ms, faster than 100.00% of C++ online submissions for Remove Nth Node From End of List.