蓝桥杯 字串分值和 如何取优化此代码的内存？

Question

#include<stdio.h>
#include<stdlib.h>
#define MAXSIZE 100000

typedef struct node{
    char data;
    struct node *next;
}linkList,LNode;

void init(linkList *L) {
    //L = (linkList*)malloc(sizeof(linkList));
    L->next = NULL;
    L->data = '\0';
}

void add(linkList *L,char x) {
    LNode *p=(LNode*)malloc(sizeof(LNode));
    p->data = x;
    p->next = L->next;
    L->next = p;
    L = p;
}


void del(linkList *L, char x) {
    LNode *p = L->next;
    int flag = 0;
    LNode * pre = L;
    while (p != NULL) {
        
        if (p->data == x&&!flag) {
            flag = 1;
            pre = p;
            p = p->next;
        }
        else if (p->data == x && flag) {
            LNode *temp = p->next;
            if (temp != NULL) {
                pre->next = temp;
                //free(temp);
            }
            else
            {
                pre->next = p->next;
            }
            p = p->next;
        }
        else {
            pre = p;
            p = p->next;
        }
    }
}

main() {
    char *S = (char*)malloc(sizeof(char)*MAXSIZE);
    scanf_s("%s", S, MAXSIZE);
    int n = 1;
    for (n = 1; S[n - 1] != '\0'; n++);
    int sum = 0;
    linkList * L=(linkList*)malloc(sizeof(linkList));
    
    
    for (int i = 0; i < n-1; i++) {
        for (int j = i; j < n-1; j++) {
            init(L);
            int count = 0;
            for (int k = i; k <= j; k++) {
                add(L, S[k]);
            }
            LNode *head = L;
            LNode *p = head->next;
            while (p!= NULL) {
                char c = p->data;
                LNode *t = p->next;
                while (t!=NULL) {
                    if (c == t->data)
                        del(L, c);
                    t = t->next;
                }
                free(t);
                p = p->next;
            }
            free(p);
            LNode *p3 = L;
            while (p3->next != NULL) {
                count++;
                p3 = p3->next;
            }
            sum += count;
        }

    }
    printf("%d", sum);
}

Answer 1

```c
int F(char *str)
{
    int check;
    int sumall;
    int sumsub;

    sumall = 0;

    for( ; *str != 0; str++)
    {
        check = 0x03FFFFFF;
        sumsub = 0;

        for(char *c = str; *c != 0; c++)
        {
            if(/*check > 0 && //为什么又封起来了，因为有26个字母的字符串不大可能*/ (((int)1 << (*c - 'a')) & check) != 0)
            {
                sumsub++;
                check = check - ((int)1 << (*c - 'a')); //修改了，再减少一次运算，不过我以前有测试过位运算比加减法要快一些的，快多少就忘记了，不知道两个位运算和一个减法到底谁快
            }

            sumall = sumall + sumsub;
        }
    }

    return sumall;
}

```