n个for循环代码实现思路

记录一道算法题:如下

根据字符序列0123456789abcdefghijklmnopqrstuvwxyz，按照序列顺序穷举生成一个N位小写字母和数字组合（N>1），每生成一个组合，不能全是字母或者全是数字。（例如用Java实现或者其它）

全身数字的情况，或者全是字母，则不行：
1111，222，01，aaaa,abcd,eb

数字和字母混合则可以：
a00,00a,aa1,ce2f

例如：
N =3
调用第一次
00a
调用第二次，输出:
00b
调用第三次，输出:
00c

当时考虑的是如果是N=2,可以把字符序列变为数组或者集合,然后遍历两个数组/集合就可以了(然后经过正则表达式处理),让给N=3,可以三个遍历,但是如果是N个的话,就没办法直接N个遍历了,

后来想到的是

(草图)

主要是for循环递归,通过递归方式，把n阶循环转变到2层for循环 

应该可以完成N个for循环,代码如下(实现思路和代码)

public static void recursionFor2() {
        Map> map = new HashMap<>();
//        String[] l1 = {"0","1","2","3","4","5","6","7","8","9","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[] l2 = {"0","1","2","3","4","5","6","7","8","9","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[] l3 = {"0","1","2","3","4","5","6","7","8","9","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[] l1 = {"0","1","2"};
        String[] l2 = {"0","1","2"};
        String[] l3 = {"0","1","2"};
        map.put(1, new HashSet<>(Arrays.asList(l1)));
        map.put(2, new HashSet<>(Arrays.asList(l2)));
        map.put(3, new HashSet<>(Arrays.asList(l3)));
        List xx = getAllStrings(map);
        System.out.println(xx.size());
        System.out.println(xx);
    }

    public static List getAllStrings(Map> map){
        if(map.size() == 2){
            List result = new ArrayList<>();
            Set tagTypeSet = map.keySet();
            Iterator iterator = tagTypeSet.iterator();
            Integer tag1 = iterator.next();
            Integer tag2 = iterator.next();
            Set set1 = map.get(tag1);
            Set set2 = map.get(tag2);
            for (String tagId1 : set1) {
                for (String tagId2 : set2) {
                    String string = new String();
                    string = tagId1+tagId2;
                    result.add(string);
                }
            }
            return result;
        }else {
            List result = new ArrayList<>();
            Set keySet = map.keySet();//获取key值
            System.out.println("tagTypeSet = " + keySet);
            Iterator iterator = keySet.iterator();//
            System.out.println("iterator = " + iterator);
            if(iterator.hasNext()){
                Integer tagValue = iterator.next();
                System.out.println("iterator.next() = " + tagValue);
                Set set = map.get(tagValue);
                Map> newMap = new HashMap<>();
                newMap.putAll(map);
                newMap.remove(tagValue);
                List oneResult = getAllStrings(newMap);
                for (String stringMap: oneResult){
                    for (String tagId : set){
                        String string = new String();
                        string = stringMap+tagId;
                        result.add(string);
                    }
                }
            }
            return result;
        }
    }

用 数组{"0","1","2"}进行验证,可以得到

[000, 001, 002, 010, 011, 012, 020, 021, 022, 100, 101, 102, 110, 111, 112, 120, 121, 122, 200, 201, 202, 210, 211, 212, 220, 221, 222]

结果正常. 

 以上方法主要是for循环递归,通过递归方式，把n阶循环转变到2层for循环

这里暂时没考虑到时间空间复杂度,后续可以继续优化