面试测试岗位常常要求手写的代码系列【一】

public class Sort {

public static void main(String[] args) {


    String str = "i-am-a-boy";
    String[] words = str.trim().split("-");

    for (int i = words.length - 1; i >= 0; i--) {
        System.out.print(words[i] + "-");
    }
    System.out.print("!!!注意多了一个-，是空格往往看不出来" + "n");


    String strb = "i-am-a-boy";
    String[] wordb = strb.trim().split("-");

    for (int i = wordb.length - 1; i >= 0; i--) {
        if (i == 0) {
            System.out.print(wordb[i]);
        } else
            System.out.print(wordb[i] + "-");
    }
    System.out.print("修复了一下，将最后一个不输出-了" + "n");

    //下面是多个空格的处理方式
    String str2 = "i---am--a-boy";
    String[] words2 = str2.trim().split("-");//会有空字符串的出现
    System.out.println("----字符串数组的长度是-----：" + words2.length);

    for (int i = words2.length - 1; i >= 0; i--) {
        if (i == 0) {
            System.out.println(words2[i]);
        } else
            System.out.print(words2[i] + "-");
    }
    //思路是变成两个字符串数组，
    // 一个是有字符串（不为空）的列表list
    // 一个是记录---的数组；这边又考虑了两种方式，一种正则表达式截取字符串（how to），一种是数-的个数，怎么数，用上述数组中到空格
    // 然后拼起来
    //String newstr = str2.replace("-"," ");//使用库函数
    String newstr2 = str2.replaceAll("[a-zA-Z]","*");//
    System.out.println(newstr2);
    String newstr3 = newstr2.replace("-","R");//使用
    System.out.println(newstr3);
    //1
    List<String> strlist = Lists.newArrayList();
    for (int i = words2.length - 1; i >= 0; i--) {
        if (words2[i].length() != 0) {//
            strlist.add(words2[i]);
        }
    }
    System.out.println(strlist);


    String[] words3 = newstr3.trim().split("*");//会有空字符串的出现
    List<String> sinlist = Lists.newArrayList();
    for (int i = 0; i <=words3.length-1; i++) {
        if (words3[i].length() != 0) {//
            sinlist.add(words3[i]);
        }
    }
    System.out.println(sinlist);
    //Collections.replaceAll(sinlist,"a","-");


    if(strlist.size()-1==sinlist.size())
    {for(int i=0;i<=sinlist.size()-1;i++)
    {System.out.print(strlist.get(i)+sinlist.get(i));}
    System.out.print(strlist.get(strlist.size()-1));
    }
    else System.out.println("那就尴尬了");

    String aaa="";
    if(strlist.size()-1==sinlist.size())
    {for(int i=0;i<=sinlist.size()-1;i++)
    {aaa=aaa+(strlist.get(i)+sinlist.get(i));}
        aaa=aaa+strlist.get(strlist.size()-1);
    }

    else System.out.println("那就尴尬了");

    aaa = aaa.replace("R","-");
    System.out.println("n"+aaa);
}

}

public class Sort {

public static void main(String[] args) {

    //下面是多个空格的处理方式
    String str2 = "i---am--a-boy";
    String[] words2 = str2.trim().split("-");//会有空字符串的出现
    System.out.println("----字符串数组的长度是-----：" + words2.length);

    for (int i = words2.length - 1; i >= 0; i--) {
        if (i == 0) {
            System.out.println(words2[i]);
        } else
            System.out.print(words2[i] + "-");
    }
    //思路是变成两个字符串数组，
    // 一个是有字符串（不为空）的列表list
    // 一个是记录---的数组；这边又考虑了两种方式，一种正则表达式替换掉字母字符（how to），一种是数-的个数，
    // 然后拼起来
    String newstr2 = str2.replaceAll("[a-zA-Z]","*");//将所有的字母字符替换成为*号
    System.out.println(newstr2);

    //1
    List<String> strlist = Lists.newArrayList();
    for (int i = words2.length - 1; i >= 0; i--) {
        if (words2[i].length() != 0) {//排除空字符串
            strlist.add(words2[i]);
        }
    }
    System.out.println(strlist);

    //2
    String[] words2a = newstr2.trim().split("*");//会有空字符串的出现
    List<String> sinlist = Lists.newArrayList();
    for (int i = 0; i <=words2a.length-1; i++) {
        if (words2a[i].length() != 0) {//
            sinlist.add(words2a[i]);
        }
    }
    System.out.println(sinlist);


    String aaa="";
    if(strlist.size()-1==sinlist.size())//连接符必须少一位啊
    {for(int i=0;i<=sinlist.size()-1;i++)
    {aaa=aaa+(strlist.get(i)+sinlist.get(i));}
        aaa=aaa+strlist.get(strlist.size()-1);
    } else System.out.println("那就尴尬了");

    System.out.println("n"+aaa);
}

}