正则表达式的匹配问题
发布时间:2020-12-14 04:21:28 所属栏目:百科 来源:网络整理
导读:这个题看似简单,但是难度不小,jdk也有实现,但是自己来造轮子发现考虑的情况非常多,这次为了这个算法思考了两三天,最后找到了解决方案如下: 我们举个例子:aaabaa 与 a*b*aa 与 a*aabaa 首先反转字符数组,因为如果不翻转,遇到了字母都要去判断下一位是
public boolean match(char[] str,char[] pattern)
{
// 反转字符数组,因为如果不翻转,遇到了字母都要去判断下一位是否是*
for (int i = str.length - 1,j = 0;j < i;i --,j ++) {
char temp = str[i];
str[i] = str[j];
str[j] = temp;
}
for (int i = pattern.length - 1,j ++) {
char temp = pattern[i];
pattern[i] = pattern[j];
pattern[j] = temp;
}
// 反转后
try {
for (int i = 0;i < pattern.length;i ++) {
if (pattern[i] == '*') {
if (pattern[i + 1] == '.')
return true;
char c = pattern[i + 1];
int t2 = i + 2;
int save2 = 0;
while (t2 < pattern.length) {
if (pattern[t2] == '*') {
t2 += 2;
} else if (pattern[t2] == c || pattern[t2] == '.') {
while (t2 < pattern.length && (pattern[t2] == c || pattern[t2] == '.')) {
save2 ++;
t2 ++;
}
break;
} else
break;
}
int p = 0;
for (char temp : str) {
if (temp == c)
p ++;
else
break;
}
if (p >= save2)
p -= save2;
// 构造新数组
char[] newArray = new char[str.length - p];
int k = 0;
while (p < str.length) {
newArray[k ++] = str[p ++];
}
str = newArray;
// 跳过一个字符串
i ++;
} else if (pattern[i] == '.') {
char[] newArray = new char[str.length - 1];
int p = 0;
for (int j = 1;j < str.length;j ++) {
newArray[p] = str[j];
}
str = newArray;
} else {
if (pattern[i] == str[0]) {
char[] newArray = new char[str.length - 1];
int p = 0;
for (int j = 1;j < str.length;j ++) {
newArray[p ++] = str[j];
}
str = newArray;
} else
return false;
}
}
} catch (Exception ex) {
return false;
}
if (str.length != 0)
return false;
return true;
}
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