LeetCode 19: Remove Nth Node From End of List

LeetCode Linked 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.
Note:
Given n will always be valid.
Try to do this in one pass.

思路

两个指针，当一个指向末尾时，另一个恰好指向要删除节点的前一个节点处。

代码

C语言版本：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
    struct ListNode* front = head;
    struct ListNode* behind = head;
    
    while ( front ){
        front = front -> next;
        if ( n-- < 0 ){
            behind = behind->next;
        }
    }
    
    if (n == 0) {
        head = head->next;
    }else {
        behind->next = behind->next->next;
    }
    return head;
}