26进制，字母表示数字，以递归算法升序输出字符串

来源：互联网发布：淘宝客有没有权重编辑：程序博客网时间：2024/06/06 05:57

转载请注明出处：http://blog.csdn.net/jssongwei/article/details/53105502

主要的工具类如下：

/** * Created by SongWei on 2016/11/9. */public class Urls {    public Urls() {}    public int bigIndex = 0; //字符数组最高位下标    public char[] urls = {'a' - 1};    //循环输出num个26进制表示的字符串    public void generUrls(int num) {        for (int index = 0; index < num; index++) {//从'a'开始一直生成num个url            if (urls[0] < 'z') {                urls[0] = (char) (urls[0] + 1);                myPrint(urls);//打印urls内容                continue;            } else {                //准备递归处理进位问题                carry2Forward(1);            }        }    }    //向前进一位    private void carry2Forward(int carry2Index) {// 只有1个元素        urls[carry2Index - 1] = 'a';        if (carry2Index <= bigIndex) {//该位未超出最大下标            if (urls[carry2Index] < 'z') {                urls[carry2Index] = (char) ((int) urls[carry2Index] + 1);                myPrint(urls);//打印urls内容            } else {                urls[carry2Index] = 'a';                carry2Forward(++carry2Index);//向高一位进位            }        } else {            carry2Head();//最高位加一个'a'            myPrint(urls);//打印urls内容        }    }    //进位到最前面，新增一最高位：urls最高位新加'a'    private void carry2Head() {        char[] newUrls = new char[bigIndex + 2];        for (int i = 0; i <= bigIndex; i++) {            newUrls[i] = urls[i];        }        bigIndex++;        newUrls[bigIndex] = 'a';        urls = newUrls;    }    //逆序打印结果（为计算方便，在char[]数组中设计成前面是低位，后面是高位）    private void myPrint(char[] chars) {        for (int i = chars.length - 1; i >= 0; i--) {            System.out.print(chars[i]);        }        System.out.print("\n");    }}

主方法如下，打印50个：

public class Main {    public static void main(String[] args) {        Urls urls = new Urls();        urls.generUrls(50);    }}

最终打印出来的结果如下：

a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
q
r
s
t
u
v
w
x
y
z
aa
ab
ac
ad
ae
af
ag
ah
ai
aj
ak
al
am
an
ao
ap
aq
ar
as
at
au
av
aw
ax
ay
az
ba
bb
bc
bd
be
bf
bg
bh
bi
bj
bk
bl
bm
bn
bo
bp
bq
br
bs
bt
bu
bv
bw
bx
by
bz
ca
cb
cc
cd
ce
cf
cg
ch
ci
cj
ck
cl
cm
cn
co
cp
cq
cr
cs
ct
cu
cv

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