标签：ora article text 简单 tar integer tutorial new 类型
做CC时经常要用正则表达式过滤数据，当时洗的数据比较复杂，规则比较多。这次做leetcode，复习一下Java的正则匹配。Leetcode 537. Complex Number Multiplication 从表示复数的字符串里把实部和虚部取出来。
http://blog.csdn.net/yin380697242/article/details/52049999
Pattern类，构造函数是私有类型，不能通过new新建，要通过Pattern.compile()获得一个匹配模式。Pattern类可以进行简单的匹配，如字符串的分割，整个字符串的匹配。
Matcher类，通过Pattern.matcher(string)获得Matcher对象。matcher.find()方法，尝试在输入字符串里进行下一次匹配，每做一次匹配后m.start()和m.end()都会改变。
Greedy/Reluctant/Possessive https://docs.oracle.com/javase/tutorial/essential/regex/quant.html
Greedy模式如X? 首先尝试匹配整个字符串，若没找到匹配，则退掉字符串最后一个字符，重新尝试匹配。
Reluctant模式如X?? 首先从字符串头开始匹配，若没找到匹配，每次加一个字符，重新匹配。
Possesive模式如X?+ 尝试匹配整个字符串，若没找到匹配，直接返回结果。
例子：
Enter your regex: .*foo // greedy quantifier Enter input string to search: xfooxxxxxxfoo I found the text "xfooxxxxxxfoo" starting at index 0 and ending at index 13. Enter your regex: .*?foo // reluctant quantifier Enter input string to search: xfooxxxxxxfoo I found the text "xfoo" starting at index 0 and ending at index 4. I found the text "xxxxxxfoo" starting at index 4 and ending at index 13. Enter your regex: .*+foo // possessive quantifier Enter input string to search: xfooxxxxxxfoo No match found.
*, ?, +的区别
*和?都可以匹配0个或多个，存在zero-length匹配，而+匹配时至少存在一个
Enter your regex: a? Enter input string to search: a I found the text "a" starting at index 0 and ending at index 1. I found the text "" starting at index 1 and ending at index 1. Enter your regex: a* Enter input string to search: a I found the text "a" starting at index 0 and ending at index 1. I found the text "" starting at index 1 and ending at index 1. Enter your regex: a+ Enter input string to search: a I found the text "a" starting at index 0 and ending at index 1.
*和? 的区别
在下面的情况中，对?匹配了多次，而*匹配了最长的一次。
Enter your regex: a? Enter input string to search: aaaaa I found the text "a" starting at index 0 and ending at index 1. I found the text "a" starting at index 1 and ending at index 2. I found the text "a" starting at index 2 and ending at index 3. I found the text "a" starting at index 3 and ending at index 4. I found the text "a" starting at index 4 and ending at index 5. I found the text "" starting at index 5 and ending at index 5. Enter your regex: a* Enter input string to search: aaaaa I found the text "aaaaa" starting at index 0 and ending at index 5. I found the text "" starting at index 5 and ending at index 5. Enter your regex: a+ Enter input string to search: aaaaa I found the text "aaaaa" starting at index 0 and ending at index 5.
这道Leetcode取实部和虚部的办法：
int[] extractOp(String complex) {
Pattern p = Pattern.compile("([-0-9]*)\\+([-0-9]*)i");
Matcher m = p.matcher(complex);
String tmp;
int[] re = new int[2];
while(m.find()) {
tmp = m.group(1);
if (tmp.startsWith("-")) {
re[0] = -(int)Integer.valueOf(tmp.substring(1,tmp.length()));
} else {
re[0] = Integer.valueOf(tmp);
}
tmp = m.group(2);
if (tmp.startsWith("-")) {
re[1] = -(int)Integer.valueOf(tmp.substring(1,tmp.length()));
} else {
re[1] = Integer.valueOf(tmp);
}
}
return re;
}
标签：ora article text 简单 tar integer tutorial new 类型
原文地址：http://www.cnblogs.com/yuchenkit/p/6789435.html
