Java 一个字符串在另外一个字符串出现次数

统计一个字符串在另外一个字符串出现次数

代码如下：

package me.chunsheng.javatest;

import java.util.regex.Matcher;
import java.util.regex.Pattern;

/**
 * Created by wei_spring on 16/10/11.
 * <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:3
 * finderTime:1579549
 * getCount:4
 * getCountTime:34506
 */
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 = 0;
        int length = source.length() - sub.length();
        for (int i = 0; i < length; i++) {
            String sourceBak = source.substring(i, i + sub.length());
            int index = sourceBak.indexOf(sub);
            if (index != -1) {
                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 = 0;
        while (matcher.find()) {
            n++;
        }
        return n;
    }
}