题解 | 找出单向链表中的一个节点，该节点到尾指针的距离为K

#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int val;
    struct ListNode* next;
};

// 创建链表
struct ListNode* createList(int len) {
    struct ListNode* testlist = (struct ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
    head = testlist;
    for (int i = 1; i <= len; i++) {
        testlist->val = i;
        if (i == len) {
            testlist->next = NULL;
            break;
        }
        testlist->next = (struct ListNode*)malloc(sizeof(struct ListNode));
        testlist = testlist->next;
    }
    return head;
}

struct ListNode* Find_k(struct ListNode* head, int k) {
    struct ListNode* temp1 = head;
    struct ListNode* temp2 = head;
    int i = 0;
    // 两个指针一前一后遍历，间隔k个节点
    while (temp1->next != NULL) {
        temp1 = temp1->next;
        i++;
        if (i >= k) {
            temp2 = temp2->next;
        }
    }
    printf("%d", temp2->val);
    return 0;
}

int main() {
    int a;
    int len = 7;
    struct ListNode* head;
    scanf("%d", &a);
    head = createList(len);
    Find_k(head, a);
    free(head);
}