bool DeleteSqlistMin(Seqlist *p,int * x )
{if(p->size==0)return false;int min=p->array[0];int i=0;int temp=0;for(i=0;i<p->size;i++){if(min>p->array[i]){min=p->array[i];temp=i;}}p->array[temp]=p->array[p->size-1];p->size--;//这里答案上写了要删除最后一项，但根据题意，本人认为无伤大雅return true;
}

bool ReverseSqlist(Seqlist *L)//由于题目只要求了空间复杂度，所以刚拿到题目就想到了用双循环冒泡法
{if(L->size==0)return false;int temp=0;int i=0;int j=0;for(i=0;i<L->size-1;i++)for(j=0;j< L->size-i-1;j++){   temp=L->array[j]; L->array[j]=L->array[j+1];L->array[j+1]=temp;}return true;
}bool ReverseSqlist2(Seqlist *L)//标答给出的折半互换法
{if(L->size==0)return false;int temp=0;int i=0;for(i=0;i<=L->size/2;i++){   temp=L->array[i]; L->array[i]=L->array[L->size-i-1];L->array[L->size-i-1]=temp;}return true;
}

bool DeleteElem(Seqlist * L,int x )
{if(L->size==0)return false;int count= 0;int distance =0;int i=0;for(i=0;i<L->size-1;i++){if(L->array[i]==x){count++;}elseL->array[i-count]=L->array[i];}L->size-=count;return true;
}

bool DelteBtwElem(Seqlist * L,int s,int t )
{int count_s=0;int count_t=0;for(int i = 0;i<L->size;i++){if(L->array[i]==s){count_s=i;break;}else if(L->array[i]!=s&&i==L->size-1){    printf("can not found s! n");return false;}}for(int i = 0;i<L->size;i++){if(L->array[i]==t&& (L->array[i+1]>L->array[i]||i==L->size-1)){count_t=i;break;}else if(L->array[i]!=t&&i==L->size-1){    printf("can not found t! n");return false;}}for(int i = count_t  ; i<L->size;i++){L->array[i- count_s -2]=L->array[i+1];}L->size-= count_t - count_s +1;return true;
}