C语言有序顺序表删除相同元素的代码，不能出来结果？

Question

#include<stdio.h>

#define MaxSize 10 typedef struct sqlist{ int data[MaxSize]; int length; }Sqlist;

void InitList(Sqlist *L){ int i; for(i=0;i<MaxSize;i++) L->data[i]=0; L->length=0; }

void keshihua(Sqlist *L){ int i; for(i=0;ilength;i++){ printf("%d ",L->data[i]); } printf("\n"); }

int main(){ void keshihua(Sqlist *L); void InitList(Sqlist *L); int j; int x,k; Sqlist *a; InitList(a);

a->data[0]=1;
a->data[1]=2;
a->data[2]=3;
a->data[3]=3;
a->data[4]=4;
a->data[5]=6;
a->data[6]=6;
a->data[7]=6;
a->data[8]=7;
a->data[9]=8;
a->length=10;

k=0;
x=a->data[0];
for(j=1;j<MaxSize;j++){
    if(x!=a->data[j]){
        a->data[j-k]=a->data[j];
        x=a->data[j];
    }
    else
        k++;
}
a->length=(a->length)-k;
keshihua(a);

}

Answer 1

你主要是没有给结构体指针分配空间

a = (Sqlist *)malloc(sizeof(Sqlist ));

代码如下：（如有帮助，望采纳！谢谢! 点击我这个回答右上方的【采纳】按钮）

#include<stdio.h>
#include <stdlib.h>
#define MaxSize 10
typedef struct sqlist{
    int data[MaxSize];
    int length;
}Sqlist;

void InitList(Sqlist *L){
    int i;
    for(i=0;i<MaxSize;i++)
        L->data[i]=0;
    L->length=0;
}

void keshihua(Sqlist *L){
    int i;
    for(i=0;i<L->length;i++){
        printf("%d ",L->data[i]);
    }
    printf("\n");
 }

int main(){
    void keshihua(Sqlist *L);
    void InitList(Sqlist *L);
    int j; int x,k;
    Sqlist *a;
    a = (Sqlist *)malloc(sizeof(Sqlist ));
    InitList(a);

    a->data[0]=1;
    a->data[1]=2;
    a->data[2]=3;
    a->data[3]=3;
    a->data[4]=4;
    a->data[5]=6;
    a->data[6]=6;
    a->data[7]=6;
    a->data[8]=7;
    a->data[9]=8;
    a->length=10;

    k=0;
    x=a->data[0];
    for(j=1;j<MaxSize;j++){
        if(x!=a->data[j]){
            a->data[j-k]=a->data[j];
            x=a->data[j];
        }
        else
            k++;
    }
    a->length=(a->length)-k;
    keshihua(a);
}

Answer 2

问题挺多的，代码给你改了，如有帮助，请帮忙采纳一下，谢谢。

#include<stdio.h>
#include <stdlib.h>
#define MaxSize 10 
typedef struct sqlist
{ 
    int data[MaxSize]; 
    int length; 
}Sqlist;

void InitList(Sqlist *L){ int i; for(i=0;i<MaxSize;i++) L->data[i]=0; L->length=0; }

void keshihua(Sqlist *L){ int i; for(i=0;i<L->length;i++){ printf("%d ",L->data[i]); } printf("\n"); }

int main(){ 
    void keshihua(Sqlist *L); 
    void InitList(Sqlist *L); 
    int i,j,t; 
    //int x,k; 
    Sqlist *a = (Sqlist*)malloc(sizeof(Sqlist));
    InitList(a);
    a->data[0]=1;
    a->data[1]=2;
    a->data[2]=3;
    a->data[3]=3;
    a->data[4]=4;
    a->data[5]=6;
    a->data[6]=6;
    a->data[7]=6;
    a->data[8]=7;
    a->data[9]=8;
    a->length=10;
    
    for(j=1;j<a->length;j++)
    {
        for (i=0;i<j;i++)
        {
            if (a->data[i] == a->data[j])
            {
                //前移
                for (t= j;t<a->length-1;t++)
                {
                    a->data[t] = a->data[t+1];
                }
                a->length -= 1;
                j--; //回退一下
            }
        }
    }
    //a->length=(a->length)-k;
    keshihua(a);
    free(a);
    return 0;
}

Answer 3

您好，我是有问必答小助手，您的问题已经有小伙伴帮您解答，感谢您对有问必答的支持与关注！
PS：问答VIP年卡 【限时加赠：IT技术图书免费领】，了解详情>>> https://vip.csdn.net/askvip?utm_source=1146287632