题解 | 删除链表的倒数第n个节点

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @param n int整型 
 * @return ListNode类
 */
#include <stdio.h>
struct ListNode* removeNthFromEnd(struct ListNode* head, int n ) {
    // write code here
    int sum = 0;
    struct ListNode* beginNode = head;
    struct ListNode* temp = head;

    while(temp){
        sum++;
        temp = temp->next;
    }

    // 处理删除头节点的情况
    if(sum == n){
        beginNode = beginNode->next;
        return beginNode;
    }

    // 移动到倒数第n个结点的前驱结点
    for(int i=0; i < sum-n-1; i++){
        head = head->next;
    }

    head->next = head->next->next;

    return beginNode;

}