Tuesday, December 1, 2015

第三十一章:带通配符的字符串匹配问题 - Art Of Programming-July


https://github.com/tdoly/The-Art-Of-Programming-by-July/blob/master/ebook/zh/31.0.md
https://github.com/julycoding/The-Art-Of-Programming-By-July

Related: Wildcard Matching, Leetcode
字符串匹配问题,给定一串字符串,按照指定规则对其进行匹配,并将匹配的结果保存至output数组中,多个匹配项用空格间隔,最后一个不需要空格。
要求:
  1. 匹配规则中包含通配符?和*,其中?表示匹配任意一个字符,*表示匹配任意多个(>=0)字符。
  2. 匹配规则要求匹配最大的字符子串,例如a*d,匹配abbdd而非abbd,即最大匹配子串。
  3. 匹配后的输入串不再进行匹配,从当前匹配后的字符串重新匹配其他字符串。
请实现函数:
    char* my_find(char input[], char rule[])
举例说明:
31.1.jpg
注意事项:
1. 自行实现函数my_find,勿在my_find函数里夹杂输出,且不准用C、C++库,和Java的String对象;
2. 请注意代码的时间,空间复杂度,及可读性,简洁性;
3. input=aaa,rule=aa时,返回一个结果aa,即可。
1. 本题与上述第三十章的题不同,上题字符串转换成整数更多考察对思维的全面性和对细节的处理,本题则更多的是编程技巧。闲不多说,直接上代码:
//copyright@cao_peng 2013/4/23  
int str_len(char *a) {  //字符串长度  
    if (a == 0) {  
        return 0;  
    }  
    char *t = a;  
    for (;*t;++t)  
        ;  
    return (int) (t - a);  
}  

void str_copy(char *a,const char *b,int len) {  //拷贝字符串 a = b  
    for (;len > 0; --len, ++b,++a) {  
        *a = *b;  
    }  
    *a = 0;  
}  

char *str_join(char *a,const char *b,int lenb) { //连接字符串 第一个字符串被回收  
    char *t;  
    if (a == 0) {  
        t = (char *) malloc(sizeof(char) * (lenb + 1));   
        str_copy(t, b, lenb);  
        return t;  
    }  
    else {  
        int lena = str_len(a);  
        t = (char *) malloc(sizeof(char) * (lena + lenb + 2));  
        str_copy(t, a, lena);  
        *(t + lena) = ' ';  
        str_copy(t + lena + 1, b, lenb);  
        free(a);  
        return t;  
    }  
}  

int canMatch(char *input, char *rule) { // 返回最长匹配长度 -1表示不匹配   
    if (*rule == 0) { //已经到rule尾端  
        return 0;  
    }  
    int r = -1 ,may;  
    if (*rule == '*') {  
        r = canMatch(input, rule + 1);  // *匹配0个字符  
        if (*input) {  
            may = canMatch(input + 1, rule);  // *匹配非0个字符  
            if ((may >= 0) && (++may > r)) {  
                r = may;  
            }  
        }  
    }  
    if (*input == 0) {  //到尾端  
        return r;  
    }  
    if ((*rule == '?') || (*rule == *input)) {  
        may = canMatch(input + 1, rule + 1);  
        if ((may >= 0) && (++may > r)) {  
            r = may;  
        }  
    }  
    return r;  
}  

char * my_find(char input[], char rule[]) {  
    int len = str_len(input);  
    int *match = (int *) malloc(sizeof(int) * len);  //input第i位最多能匹配多少位 匹配不上是-1  
    int i,max_pos = - 1;  
    char *output = 0;  

    for (i = 0; i < len; ++i) {  
        match[i] = canMatch(input + i, rule);  
        if ((max_pos < 0) || (match[i] > match[max_pos])) {  
            max_pos = i;  
        }  
    }  
    if ((max_pos < 0) || (match[max_pos] <= 0)) {  //不匹配  
        output = (char *) malloc(sizeof(char));  
        *output = 0;   // \0  
        return output;  
    }  
    for (i = 0; i < len;) {  
        if (match[i] == match[max_pos]) { //找到匹配  
            output = str_join(output, input + i, match[i]);  
            i += match[i];  
        }  
        else {  
            ++i;  
        }  
    }  
    free(match);  
    return output;  
}  
2. 本题也可以直接写出DP(Dynamic Programming, 动态规划)方程,如下代码所示:
//copyright@chpeih 2013/4/23  
char* my_find(char input[], char rule[])  
{  
    //write your code here  
    int len1, len2;  
    for(len1 = 0; input[len1]; len1++);  
    for(len2 = 0; rule[len2]; len2++);  
    int MAXN = len1 > len2 ? (len1+1) : (len2+1);  
    int  **dp;  

    //dp[i][j]表示字符串1和字符串2分别以i j结尾匹配的最大长度  
    //记录dp[i][j]是由之前那个节点推算过来  i*MAXN+j  
    dp = new int *[len1+1];  
    for (int i = 0;i<=len1;i++)  
    {  
        dp[i] = new int[len2+1];  

    }  

    dp[0][0] = 0;  
    for(int i = 1; i <= len2; i++)  
        dp[0][i] = -1;  
    for(int i = 1; i <= len1; i++)  
        dp[i][0] = 0;  

    for (int i = 1; i <= len1; i++)  
    {  
        for (int j = 1; j <= len2; j++)  
        {  
            if(rule[j-1] == '*'){  
                dp[i][j] = -1;  
                if (dp[i-1][j-1] != -1)  
                {  
                    dp[i][j] = dp[i-1][j-1] + 1;  

                }  
                if (dp[i-1][j] != -1 && dp[i][j] < dp[i-1][j] + 1)  
                {  
                    dp[i][j] = dp[i-1][j] + 1;  

                }  
            }else if (rule[j-1] == '?')  
            {  
                if(dp[i-1][j-1] != -1){  
                    dp[i][j] = dp[i-1][j-1] + 1;  

                }else dp[i][j] = -1;  
            }   
            else  
            {  
                if(dp[i-1][j-1] != -1 && input[i-1] == rule[j-1]){  
                    dp[i][j] = dp[i-1][j-1] + 1;  
                }else dp[i][j] = -1;  
            }  
        }  
    }  

    int m = -1;//记录最大字符串长度  
    int *ans = new int[len1];  
    int count_ans = 0;//记录答案个数  
    char *returnans = new char[len1+1];  
    int count = 0;  
    for(int i = 1; i <= len1; i++)  
        if (dp[i][len2] > m){  
            m = dp[i][len2];  
            count_ans = 0;  
            ans[count_ans++] = i-m;  
        }else if(dp[i][len2] != -1 && dp[i][len2] == m){  
            ans[count_ans++] = i-m;  
        }  

        if (count_ans!=0)  
        {      
            int len = ans[0];  
            for (int i = 0;i < m;i++)  
            {  
                printf("%c",input[i+ans[0]]);  
                returnans[count++] = input[i+ans[0]];  
            }  
            for (int j = 1;j<count_ans;j++)  
            {  
                printf(" ");  
                returnans[count++] = ' ';  
                len = ans[j];  
                for (int i = 0;i<m;i++)  
                {  
                    printf("%c",input[i+ans[j]]);  
                    returnans[count++] = input[i+ans[j]];  
                }  
            }  
            printf("\n");  
            returnans[count++] = '\0';  
        }  

        return returnans;  
}  


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