算法导论习题32.1-4解答

从一个参考书上得到的求解算法，描述如下：

To determine if a pattern P with gap characters exists in T partition P into substrings P1, ..., Pk determined by the gap characters. Search for P1 and if found continue searching for P2 and so on. This clearly find a pattern if one exists.

由以上思路不难写出代码，如下：

#include <iostream>#include <cstring>using namespace std;//使用朴素的字符串比较算法，返回第一个匹配的起始地点，若没有则为NULLconst char* findFirstContainMatch(const char* text, const char* pattern){    int textLen = strlen(text);    int patternLen = strlen(pattern);    for(int s = 0; s <= textLen - patternLen; ++s){        int i =0;        for(; i < patternLen; ++i){            if(text[s+i] != pattern[i]){                break;            }        }        //包含        if(i == patternLen){            return (text + s);        }    }    return NULL;}bool isSubStr(const char* text, const char* pattern){    while(true){        //找到某个不包含分隔符的子串的开始处        for(; *pattern == '*'; ++pattern);        cout<<*pattern <<" ";        if(*pattern == '\0'){            return true;        }        //找到第一个相匹配的地方        for(; *text != '\0' && *text != *pattern; ++text);        if(*text == '\0'){            return false;        }        //统计当前模式中下一个不包含分隔符的子串的长度        unsigned patternLen = 0;        for(const char* temp = pattern; *temp != '\0' && *temp != '*'; ++temp){            ++patternLen;        }        //将模式中某个不包含分隔符的子串提取出来        char* subStr = new char[patternLen+1];        subStr[patternLen] = '\0';        strncpy(subStr, pattern, patternLen);        if((text = findFirstContainMatch(text, subStr)) == NULL){            return false;        }        //开始匹配下一个子串        pattern = pattern + patternLen;        text = text + patternLen;        delete[] subStr;    }    return false;}int main(){    char text[] = "cabccbacbacba";    char pattern[] = "****abcc*b*a*c*b*b*a";    cout << (isSubStr(text, pattern) ? "find it!" : "not find!") <<endl;    return 0;}