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 quantifierEnter input string to search: xfooxxxxxxfooI found the text "xfooxxxxxxfoo" starting at index 0 and ending at index 13.Enter your regex: .*?foo  // reluctant quantifierEnter input string to search: xfooxxxxxxfooI 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 quantifierEnter input string to search: xfooxxxxxxfooNo 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: aI 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: aI 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: aI 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: aaaaaI 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: aaaaaI 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: aaaaaI 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;    }