【链表】4、带头结点的单链表删除最小值节点

/*
    删除链表中唯一的最小值
    分析：
        目前能想到的就是遍历整个链表，记录下最小节点的指针，然后进行删除，时间复杂度为O(n).
*/
struct Link {
    int data;
    struct Link* next;
};
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
void deleteMin(Link *p) {
    struct Link *preMinp = p, *minP = p->next,*preQ=p->next, *q = p->next->next,*f;
    while (q) {
        if (q->data < minP->data) {//如果比当前值更小
            minP = q;//更换
            preMinp = preQ;//前驱一起更换
        }
        preQ = q;//继续前进
        q = q->next;
    }
    f = minP;
    preMinp->next = minP->next;//删除
    free(f);//释放
}
int main() {
        int n,data;
        printf("请输入创建链表的节点个数：");
        scanf("%d",&n);
        struct Link *q;
        struct Link *head =(struct Link*) malloc(sizeof(struct Link));
        head->next = NULL;
        q = head;
        for (int i = 0; i < n;i++) {
            struct Link *newP = (struct Link*) malloc(sizeof(struct Link));
            printf("请输入第一个节点的值：");
            scanf("%d",&data);
            newP->data = data;
            newP->next = NULL;
            head->next = newP;
            head = head->next;//head要始终指向最新节点
        }
        head->next = NULL;
        head = q;//最后head要指向头结点
        printf("打印链表：");
        while (head->next) {
            printf("%d ", head->next->data);
            head = head->next;
        }
        head = q;
        deleteMin(head);
        printf("删除后链表值为：");
        while (head->next) {
            printf("%d ", head->next->data);
            head = head->next;
        }
        return 0;
}