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   本题代码如下（有注释）

void delete_min(linklist* head)
{while ((*head)->next != NULL)//循环到只剩下头节点{lnode* pre = *head;//pre为元素最小结点的前驱结点指针lnode* p = (*head)->next;//p为工作指针lnode* q;//指向被删除的结点while (p->next != NULL)//找到最小的结点{if (p->next->data < pre->next->data)pre = p;//记住当前最小值的前驱p = p->next;}printf("%d ", pre->next->data);//输出最小值结点的数据q = pre->next;//p指向最小结点pre->next = q->next;//pre后继指向q的后继free(q);//释放q结点}free(*head);//释放头节点
}

完整测试代码

#include<stdio.h>
#include<stdlib.h>
typedef struct lnode
{int data;struct lnode* next;
}lnode,*linklist;
int n = 5;
int a[5] = { 2,1,9,8,7 };
void buildlinklist(linklist* head)//建立单链表
{*head = (lnode*)malloc(sizeof(lnode));(*head)->next = NULL;int i = 0;lnode* s = *head, * r = *head;for (i = 0; i < n; i++){s = (lnode*)malloc(sizeof(lnode));s->data = a[i];s->next = r->next;r->next = s;r = s;}r->next = NULL;
}
void delete_min(linklist* head)
{while ((*head)->next != NULL)//循环到只剩下头节点{lnode* pre = *head;//pre为元素最小结点的前驱结点指针lnode* p = (*head)->next;//p为工作指针lnode* q;//指向被删除的结点while (p->next != NULL)//找到最小的结点{if (p->next->data < pre->next->data)pre = p;//记住当前最小值的前驱p = p->next;}printf("%d ", pre->next->data);//输出最小值结点的数据q = pre->next;//p指向最小结点pre->next = q->next;//pre后继指向q的后继free(q);//释放q结点}free(*head);//释放头节点
}
int main()
{linklist head;buildlinklist(&head);delete_min(&head);return 0;
}