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jincc
iOS-Algorithm
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//
// Pattern.h
// backtracking
//
// Created by junl on 2019/9/12.
// Copyright © 2019 junl. All rights reserved.
//
#ifndef Pattern_hpp
#define Pattern_hpp
#include <stdio.h>
#include <string.h>
/*
正则表达式匹配,假设通配符只有两个,一个是*,一个是?
* 代表可以匹配任意多个(>=0)任意字符.
? 代表匹配另个或一个任意字符
对于字符串的匹配,对于字符来说有两种情况:
1.如果是非通配符情况下,必须完全匹配
2.如果是通配字符,就有很多种选择了,就拿*来说,它可以匹配任意多个任意字符,这里我们利用回溯的思想穷举出所有的情况.
*/
typedef struct {
const char *str;
size_t slen;
const char *pattern;
size_t plen;
bool match;
}match_t;
void match(match_t *context, int pidx, int sidx);
bool isMatch(const char *str, const char *pattern){
match_t m = (match_t){str, strlen(str), pattern, strlen(pattern), false};
match(&m, 0, 0);
return m.match;
}
void match(match_t *context, int pidx, int sidx){
if (context->match) return;
if (pidx == context->plen){
if (sidx == context->slen)
context->match = true;
}
char ch = context->pattern[pidx];
if (ch == '?'){
match(context, pidx+1, sidx);
match(context, pidx+1, sidx+1);
}else if (ch == '*'){
for (int i=0; i <= context->slen - sidx; i++) {
match(context, pidx+1, sidx+i);
}
}else if (sidx < context->slen && ch == context->str[sidx]){
match(context, pidx+1, sidx+1);
}
}
#endif /* Pattern_hpp */
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