任意没有。公差等级。

为我能想到的所有测试用例工作。松散地基于| / | ads解决方案。

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

report (int x, char* str, char* sstr, int[] t) {
    if ( x )
        printf( "%s is a substring of %s for a tolerance[%d]\n",sstr,str[i],t[i] );
    else
        printf ( "%s is NOT a substring of %s for a tolerance[%d]\n",sstr,str[i],t[i] );
}

int find_with_tolerance (char *str, char *sstr, int tol) {

    if ( (*sstr) == '\0' ) //end of substring, and match
        return 1;

    if ( (*str) == '\0' ) //end of string
        if ( tol >= strlen(sstr) ) //but tol saves the day
            return 1;
        else    //there's nothing even the poor tol can do
            return 0;

    if ( *sstr == *str ) { //current char match, smooth
        return find_with_tolerance ( str+1, sstr+1, tol );
    } else {
        if ( tol <= 0 ) //that's it. no more patience
            return 0;
        for(int i=1; i<=tol; i++) {
            if ( *(str+i) == *sstr ) //insertioan of a foreign character
                return find_with_tolerance ( str+i+1, sstr+1, tol-i );
            if ( *str == *(sstr+i) ) //deal with dletion
                return find_with_tolerance ( str+1, sstr+i+1, tol-i );
            if ( *(str+i) == *(sstr+i)  ) //deal with riplacement
                return find_with_tolerance ( str+i+1, sstr+i+1, tol-i );
            if ( *(sstr+i) == '\0' ) //substr ends, thanks to tol & this loop
                return 1;
        }
        return 0; //when all fails
    }
}

int find (char *str, char *sstr, int tol ) {
    int w = 0;
    while (*str!='\0')
        w |= find_with_tolerance ( str++, sstr, tol );
    return (w) ? 1 : 0;
}

int main() {
    const int n=3; //no of test cases
    char *sstr = "dog"; //the substr
    char *str[n] = { "doox", //those cases
                    "xxxxxd",
                    "xxdogxx" };
    int t[] = {1,1,0}; //tolerance levels for those cases
    for(int i = 0; i < n; i++) {
        report( find ( *(str+i), sstr, t[i] ), *(str+i), sstr, t[i] );
    }
    return 0;
}

5 个答案:

任意没有。公差等级。