【LeetCode】19. 删除链表的倒数第N个节点(Remov

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19. Remove Nth Node From End of List

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?

中文：

给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。

示例：

给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后，链表变为 1->2->3->5.

说明：

给定的 n 保证是有效的。

Accept by C:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* removeNthFromEnd(struct ListNode* head, int n){
    //设置工作指针指向头节点
    struct ListNode* p = head; 
    //获得链表长度
    int count = 0;
    while(p != NULL){
        ++count;
        p = p -> next;
    }
    //获取第length - n个节点的地址
    p = head;
    int i = 1;
    while(i++ < count - n){
        p = p -> next;
    }
    //如果要删除的是头节点
    if(count == n){
        struct ListNode* s = head;
        head = head -> next;
        free(s); //释放
        return head;
    }
    //其余情况
    struct ListNode* p1 = p -> next;
    p -> next = p -> next ->next;
    free(p1);//释放
    return head;
}

AC