Question

C# 怎样用a-z生成8位字符串的所有组合

一个字符串里的字符是可以重复的，比如aabbcdef这样子的。我想要能列出所有的组合，但是想出来的都列不全

Answer 1

char ba[]={"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"};
string s;
for(int a=0;a<ba.length;a++){
for(int b=0;b<ba.length;b++){
for(int c=0;c<ba.length;c++){
for(int d=0;d<ba.length;d++){
for(int e=0;e<ba.length;e++){
for(int f=0;f<ba.length;f++){
for(int g=0;g<ba.length;g++){
for(int h=0;h<ba.length;h++){
s=s+ba[a]+ba[b]+ba[c]+ba[d]+ba[e]+ba[f]+ba[g]+ba[h]+"\n";
}
}
}
}
}
}
}
}
代码应该就是这样的吧!C#不常用.

Answer 2

8层for循环，不过运行时很卡，要计算26的8次方次

Answer 3

先给你个思路：
例如：a,b,c,d,e,f中组合成长度为3的字符串：
1.
aaa
aab
aac
aad
aae
aaf
aba
abb
abc
abd
abe
abf
aca
acb
acc
acd
ace
acf
...
类推下去，找到规律，然后用循环穷举就可以了

Answer 4

这个数量相当庞大吧 你Java跑几天就跑不完

Answer 5

组合啥要求？

追问

长度为8，由a-z组成