Input: str[] = { “aabbccdaabbcc” }
Output: 6
Input: abdab
Output: 2
如果我们从字符串的开头和结尾开始指针,那么它们会在某个点重叠,所以我们不会这样做,而是从中间断开字符串并开始匹配左右字符串。如果它们相等,则任何一个匹配字符串的返回大小相同,否则尝试两侧的长度较短。
int longest(char str[], int n)
START
STEP 1 : DECLARE length AS 0 AND i AS n/2
STEP 2 : IF n < 2 THEN
RETURN 1
STEP 3 :LOOP WHILE TILL str[i]!=''
IF str[i] == str[length] THEN,
INCREMENT length BY 1
INCREMENT i BY 1
ELSE
IF length == 0 THEN,
INCREMENT i BY 1
ELSE
DECREMENT length BY 1
END IF
END IF
END WHILE
RETURN length
STOP
#include
int longest(char str[], int n){
int length = 0, i = n/2;
if( n < 2 )
return 1;
while( str[i]!='' ){
//When we find the character like prefix in suffix,
//we will move the length and i to count the length of the similar prefix and suffix
if (str[i] == str[length]){
++length;
++i;
} else //When prefix and suffix not equal{
if(length == 0)
++i;
else
–length;
}
}
return length;
}
int main(int argc, char const *argv[]){
char str[] = {"abccmmabcc"};
int n = sizeof(str)/sizeof(str[0]);
int length = longest(str, n);
printf("Length = %d", length);
return 0;
}
如果我们运行上面的程序,它将生成以下输出:
Length = 4
如果您发现该资源为电子书等存在侵权的资源或对该资源描述不正确等,可点击“私信”按钮向作者进行反馈;如作者无回复可进行平台仲裁,我们会在第一时间进行处理!
加入交流群
请使用微信扫一扫!