已知一个单向链表的头，请写出删除其某一个结点的算法，要求，先找到此 结点，然后删除。用C语言实现

以下是单向链表删除某一结点的C语言实现：

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

// 定义链表结点结构体
struct ListNode {
    int val;
    struct ListNode *next;
};

// 删除指定结点
void deleteNode(struct ListNode* head, int val) {
    struct ListNode* prev = head;
    struct ListNode* curr = head->next;
    while(curr != NULL) {
        if(curr->val == val) {
            prev->next = curr->next;
            free(curr);
            return;
        }
        prev = curr;
        curr = curr->next;
    }
}

int main() {
    // 创建链表
    struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
    head->val = 0;
    head->next = NULL;

    struct ListNode* node1 = (struct ListNode*)malloc(sizeof(struct ListNode));
    node1->val = 1;
    node1->next = NULL;

    struct ListNode* node2 = (struct ListNode*)malloc(sizeof(struct ListNode));
    node2->val = 2;
    node2->next = NULL;

    head->next = node1;
    node1->next = node2;

    // 删除结点2
    deleteNode(head, 2);

    // 打印链表
    struct ListNode* curr = head->next;
    while(curr != NULL) {
        printf("%d ", curr->val);
        curr = curr->next;
    }

    return 0;
}

注：此实现中假设链表中不会有重复元素。如果链表中有重复元素，需要改写删除结点的判断条件。