字符串模式匹配和插入C ++

时间:2015-08-08 23:41:34

标签: c++ algorithm matching

我正在尝试匹配并在字符串中插入一个patern。

此处Good peo Good peo,我正在搜索peo并插入ple

但输出就是这样:

Good people Good peo /n
Good peo Good people

我需要输出像

Good people Good people

我的代码:

#include<stdio.h>
#include<stdlib.h>

#include<iostream>
#include<string>
#include<string.h>
using namespace std;

int length(char s[])
{
    int len=0;
    int i=0;
    while(s[i]!='\0')
    {
        i++;
        len++;
    }
    return len;
}

void concatenate(char s1[], char s2[])
{
    int i=length(s1);
    int j=length(s2);
    int count=0;
    while(count<=j)
    {
        s1[i]=s2[count];
        i++;
        count++;
    }
}

void substring(char s[], char dest[], int ip, int len)
{
    int i=ip;
    int count=0;
    while(count<len)
    {
        dest[count]=s[i];
        count++;
        i++;
    }
    dest[count]='\0';
}

void ins(char T[], int ip, char P[])
{
    char temp1[100];
    char temp2[100];
    substring(T, temp1, 0, ip);
    substring(T, temp2, ip, length(T)-ip);
    concatenate(temp1, P);
    concatenate(temp1, temp2);
    T=temp1;
    cout<<T<<endl;
}

void del(char T[], int ip, int L)
{
    char temp1[100];
    char temp2[100];
    substring(T, temp1, 0, ip);
    substring(T, temp2, ip+L, length(T)-ip-L);
    concatenate(temp1, temp2);
    T=temp1;
    cout<<T<<endl;
}

//where T is the original string and P is the pattern to be deleted whereever it appears in the original string.
void delpat(char T[], char P[])
{
    char temp[100];
    for(int i=0; i<=length(T); i++)
    {
        substring(T, temp, i, length(P));
        if(strcmp(temp, P)==0)
            del(T, i, length(P));
    }
}

//where T is the original string, Q is the pattern to be inserted and P is the pattern after which it is inserted.
void inspat(char T[], char P[], char S[])
{
    char temp[100];
    for(int i=0; i<=length(T); i++)
    {
        substring(T, temp, i, length(P));
        if(strcmp(temp, P)==0)
            ins(T, i+length(P), S);
    }
}

int main()
{ char a[100];
    char T[]="Good peo Good peo";
    char P[]="peo";
    char S[]="ple";
    inspat(T, P, S);
    gets(a);
}

1 个答案:

答案 0 :(得分:2)

1)函数ins()中的赋值不会更改调用者的值:

T=temp1;
cout<<T<<endl;

您需要使用strcpy()复制temp1 char数组:

strcpy(T, temp1);
cout<<T<<endl;

2)由于您要打印插入所有次后的内容,因此需要执行上述cout,您可以在T中打印main()或在inspat()for循环之外):

cout<<T<<endl;

3)由于插入发生在原始数组中,因此需要确保数组足够大。做main()

之类的事情
char T[256]="Good peo Good peo"; // 256 is some arbitrary size
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