package me.chunsheng.javatest;

import java.util.regex.Matcher;

import java.util.regex.Pattern;

/**

 * Created by wei_spring on //.

 * <p>

 * 统计一个字符串在另外一个字符串出现的次数

 * 正则匹配的方法,前提字符串不是特殊字符串

 * eg:finder("adadadadauuada", "ada") 这样会统计不准确的.

 * 另外一个遍历的解决了上面那种情况.

 * 两者时间如下(纳秒为单位): 正则明显耗时,正则本就不是统计用的!!!

 * finder:

 * finderTime:

 * getCount:

 * getCountTime:

 */

public class SubStringCount {

    public static void main(String[] args) {

        long current1 = System.nanoTime();

        System.out.println("finder:" + finder("adadadadauuada", "ada"));

        System.out.println("finderTime:" + (System.nanoTime() - current1));

        long current2 = System.nanoTime();

        System.out.println("getCount:" + getCount("adadadadauuada", "ada"));

        System.out.println("getCountTime:" + (System.nanoTime() - current2));

    }

    public static int getCount(String source, String sub) {

        int count = ;

        int length = source.length() - sub.length();

        for (int i = ; i < length; i++) {

            String sourceBak = source.substring(i, i + sub.length());

            int index = sourceBak.indexOf(sub);

            if (index != -) {

                count++;

            }

        }

        return count;

    }

    public static int finder(String source, String regexStr) {

        String regex = "[a-zA-Z]+";

        if (regexStr != null && !regexStr.equals("")) {

            regex = regexStr;

        }

        Pattern expression = Pattern.compile(regex);

        Matcher matcher = expression.matcher(source);

        int n = ;

        while (matcher.find()) {

            n++;

        }

        return n;

    }

}