Question

我试图找到2个字符串之间的最大重叠并打印2个字符串，重叠只在新创建的字符串中出现一次。

#include<stdio.h>
#include<string.h>
int main(void)
{
    char string1[64],string2[64];
    char s3[128];
    int len=0;

    printf("Enter string1");
    scanf("%s",string1);
    printf("Enter string2");
    scanf("%s",string2);
    for (char* s1 = string1; *s1 != '\0'; ++s1)
    {
        int overlap = 0;
        for (char* s2 = string2; *s2 != '\0'; ++s2)
        {
            while (s1[overlap] != '\0' && s1[overlap] == s2[overlap])
            {
                ++overlap;
            }
            if ( len < overlap )
            {
                len = overlap;
            }
        }
        s3[overlap]=string1[overlap];
    }
    printf("%d\n",len);
    printf("%s\n",s3);
    printf(strcat(string1,string2));
    return 0;
}

我找到了最大重叠和重叠中的那些字符，但我不知道如何从2个字符串的组合中减去重叠来得到我的答案。示例：a123和a123457应该给我4和a123457（a123只出现一次）。

Answer 1

事实证明这比我预期的要多。我将你的指针转换为整数索引，以便可以直接比较两个字符串（构造新字符串所必需的。）然后我添加了变量来跟踪每个字符串中最长重叠的开始：

#include <stdio.h>
#include <string.h>

#define MAX_STR_LEN (64)

int main(void) {
    char string1[MAX_STR_LEN], string2[MAX_STR_LEN], string3[MAX_STR_LEN * 2 - 1];
    int longest_overlap = 0, overlap_offset_s1 = -1, overlap_offset_s2 = -1;

    printf("Enter string1: ");
    scanf("%s", string1);

    printf("Enter string2: ");
    scanf("%s", string2);

    int s1_len = strlen(string1), s2_len = strlen(string2);

    for (int s1 = 0; s1 < s1_len; s1++)
    {
        for (int s2 = 0; s2 < s2_len; s2++)
        {
            int overlap = 0;

            while (s1 + overlap < s1_len && s2 + overlap < s2_len && string1[s1 + overlap] == string2[s2 + overlap])
            {
                ++overlap;
            }

            if (longest_overlap < overlap)
            {
                longest_overlap = overlap;
                overlap_offset_s1 = s1;
                overlap_offset_s2 = s2;
            }
        }
    }

    printf("%d\n", longest_overlap);

    strncpy(string3, &string1[overlap_offset_s1], longest_overlap);
    string3[longest_overlap] = '\0';

    printf("%s\n", string3);

    if (longest_overlap == 0)
    {
        strcat(strcpy(string3, string1), string2);
    }
    else if (overlap_offset_s1 > overlap_offset_s2 || (overlap_offset_s1 == overlap_offset_s2 && s1_len < s2_len))
    {
        strcpy(string3, string1);
        string3[overlap_offset_s1] = '\0';
        strcat(string3, &string2[overlap_offset_s2]);
    }
    else
    {
        strcpy(string3, string2);
        string3[overlap_offset_s2] = '\0';
        strcat(string3, &string1[overlap_offset_s1]);

    }

    printf("%s\n", string3);

    return 0;
}

<强>实施例

> ./a.out
Enter string1: a123
Enter string2: a123457
4
a123
a123457
> 
> ./a.out
Enter string1: abcdef
Enter string2: cdefgh
4
cdef
abcdefgh
>

> ./a.out
Enter string1: cdefgh
Enter string2: abcdef
4
cdef
abcdefgh
>

其他示例

> ./a.out
Enter string1: abc
Enter string2: 123
0

abc123
>

在新字符串中仅打印重叠的字符一次（字符串组合）

1 个答案: