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标题:.表达式问题(后缀表达式转换为中缀表达)
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漩涡鸣人——
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回复 9楼 embed_xuel
有关系吗?这要的是多练吧,如果你一个月做几百道题,你也可以的,但是你能吗?
2012-04-20 15:53
embed_xuel
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回复 11楼 漩涡鸣人——
我不能,无知者无畏。光做题就能拿一等奖,还在没学数据结构的情况下

[ 本帖最后由 embed_xuel 于 2012-4-20 16:14 编辑 ]

总有那身价贱的人给作业贴回复完整的代码
2012-04-20 16:12
jxlcs
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模拟计算后缀表达式的方式。
遇到符号则从栈中取出两个表达式或是数字进行重组成新的表达式,栈中存放当前符号计算出来的表达式

struct Node
{
    char str[50];
    int level;
    PtrToNode next;
};

//Push 进栈
void Push(char *x, int level, Stack S)
{
    PtrToNode tmpCell;
    int i=0;

    tmpCell = malloc(sizeof(struct Node));
    if (tmpCell == NULL)
        printf("Out of space!\n");
    else
    {
        strcpy(tmpCell->str, x);
        tmpCell->level = level;
        tmpCell->next = S->next;
        S->next = tmpCell;
    }
}

//从栈弹出元素
void Pop(Stack S)
{
    PtrToNode FirstCell;

    if(IsEmpty(S))
        printf("Empty stack\n");
    else
    {
        FirstCell = S->next;
        S->next = S->next->next;
        free(FirstCell);
    }
}

//返回栈顶元素所在结点的指针
Stack Top(Stack S)
{
    if( !IsEmpty(S) )
        return S->next;
    printf("Empty stack\n");
    return NULL;
}

//使栈为空
void MakeEmpty(Stack S)
{
    if (S == NULL)
        printf("Must Use CreeateStack first.\n");
    else
        while(!IsEmpty(S))
            Pop(S);
}

//创建一个空栈
Stack CreateStack(void)
{
    Stack S;
   
    S = malloc(sizeof(struct Node));
    if(S == NULL)
        printf("Out of space!\n");
    S->next = NULL;
    MakeEmpty(S);
    return S;
}

//测试栈是否空栈
int IsEmpty(Stack S)
{
    return S->next == NULL;
}

//后缀表达式转成中缀表达式
void PostfixToInfix(char *In, char *Post)
{
    int i,j,level;
    char c,s[20],s2[20], exp[30];
    Stack S,tmp;
    S = CreateStack();
    MakeEmpty(S);
    c = *Post++;
    while (c != '\0')
    {
        if (c >= '0' && c <= '9' || c == '.')
        {
            level = 0;
            i=0;
            while (c >= '0' && c <= '9' || c == '.')
            {
                s[i++] = c;
                c = *Post++;
            }
            s[i] = '\0';
            //数字字符串进栈
            Push(s, level, S);
        }
        //符号处理
        if (c == '+' || c == '-')
        {
            level = 1;
            tmp = Top(S);
            
            //如果栈顶优先级低,保存此表达式,并弹出
            if (tmp->level <= level)
            {
                i=0;    j=0;
                while (tmp->str[j]!='\0')
                    s[i++] = tmp->str[j++];
                s[i] = '\0';
                Pop(S);
            }

            tmp = Top(S);
            if (tmp->level <= level)
            {
                i=0;    j=0;
                while (tmp->str[j]!='\0')
                    s2[i++] = tmp->str[j++];
                s2[i] = '\0';
                Pop(S);
            }

            //用此运算符组成表达式
            i=0;    j=0;
            while(s2[j] != '\0')
                exp[i++] = s2[j++];
            exp[i++] = c;
            j=0;
            while(s[j] != '\0')
                exp[i++] = s[j++];
            exp[i] = '\0';

            //表达式进栈
            Push(exp, level, S);
        }
        if (c == '*' || c == '/' || c == '%')
        {
            level = 3;
            tmp = Top(S);
            //printf("top1: %s\n",tmp->str);
            //printf("top2: %s\n",tmp->next->str);
            //如果栈顶表达式优先级低,加括号并保存
            if (tmp->level == 1 || tmp->level == 2)
            {
                i=0;    j=0;
                s[i++] = '(';
                while (tmp->str[j]!='\0')
                    s[i++] = tmp->str[j++];
                s[i++] = ')';
                s[i] = '\0';
                //带括号的表达式的优先级置为 -1
                tmp->level = -1;
                //当前表达式整体优先级置为2
                level = 2;
                Pop(S);
            }else
            if (tmp->level <= level || tmp->level == -1)
            {
                i=0;    j=0;
                while (tmp->str[j]!='\0')
                    s[i++] = tmp->str[j++];
                s[i] = '\0';
                Pop(S);
            }

            tmp = Top(S);
            //如果栈顶表达式优先级低,加括号并保存
            if (tmp->level ==1)
            {
                i=0;    j=0;
                s2[i++] = '(';
                while (tmp->str[j]!='\0')
                    s2[i++] = tmp->str[j++];
                s2[i++] = ')';
                s2[i] = '\0';
                Pop(S);
            }else
            if (tmp->level <= level || tmp->level == -1)
            {
                i=0;    j=0;
                while (tmp->str[j]!='\0')
                    s2[i++] = tmp->str[j++];
                s2[i] = '\0';
                Pop(S);
            }

            //用此运算符组成表达式
            i=0;    j=0;
            while(s2[j] != '\0')
                exp[i++] = s2[j++];
            exp[i++] = c;
            j=0;
            while(s[j] != '\0')
                exp[i++] = s[j++];
            exp[i] = '\0';
            //表达式进栈
            Push(exp, level, S);
        }
        if (c == '^')
        {
            level = 0;
            tmp = Top(S);
            
            //如果栈顶优先级低,保存此表达式,并弹出
            if (tmp->level <= level)
            {
                i=0;    j=0;
                while (tmp->str[j]!='\0')
                    s[i++] = tmp->str[j++];
                s[i] = '\0';
                Pop(S);
            }

            tmp = Top(S);
            if (tmp->level <= level)
            {
                i=0;    j=0;
                while (tmp->str[j]!='\0')
                    s2[i++] = tmp->str[j++];
                s2[i] = '\0';
                Pop(S);
            }

            //用此运算符组成表达式
            i=0;    j=0;
            while(s2[j] != '\0')
                exp[i++] = s2[j++];
            exp[i++] = c;
            j=0;
            while(s[j] != '\0')
                exp[i++] = s[j++];
            exp[i] = '\0';

            //表达式进栈
            Push(exp, level, S);
        }

        c = *Post++;
        //putchar(c);
    }
    tmp = Top(S);
    strcpy(In, tmp->str);
    Pop(S);
}
2012-12-17 05:19
快速回复:.表达式问题(后缀表达式转换为中缀表达)
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