如何输出字符串的所有组合？

数据结构与算法

题目描述：假如字符串中的字符都不重复，如何输出字符串的所有组合？例如“abc”
---->a,b,c,ab,bc,ac,abc

1.首发--老办法

其实就是深搜的过程中的所有情况，。注意去重

还有一点就是，。，。这个set莫名奇妙被套了一层，后面再来重构

public class Solution {
    private Set<String> resultSet = new HashSet<>();
    public void printAllCombination(String string) {
        if (string == null || string.length() == 0) {
            return;
        }
        char[] chars = string.toCharArray();
        StringBuffer sb = new StringBuffer();
        int step = 0;
        int[] book = new int[string.length()];
        dfs(chars, sb, book, step);
        System.out.println(resultSet);
        for (String str : resultSet) {
            if (str.equals("[]")) {
                continue;
            } else {
                System.out.println("result:" + str.substring(1, str.length()-1));
            }
        }
    }
    public void dfs(char[] chars, StringBuffer sb, int[] book, int step) {
        char[] resultChar = sb.toString().toCharArray();
        Arrays.sort(resultChar);
        String resultString = Arrays.toString(resultChar);
        if (resultString != null && resultString.length() != 0) {
            resultSet.add(resultString);
        }
        if (step == book.length) {
            return;
        }
        for (int i = 0; i < chars.length; i++) {
            if (book[i] == 0) {
                book[i] = 1;
                sb.append(chars[i]);
                dfs(chars, sb, book, step + 1);
                book[i] = 0;
                sb.deleteCharAt(sb.length() - 1);
            }
        }
    }
    public static void main(String[] args) {
        Solution solution = new Solution();
        String string = "abc";
        solution.printAllCombination(string);
    }
}

2.，。，有玄妙

根据排列组合知识，，结果应为$2^n-1$个元素，。，恰好对应
1，10，11，100，。，。
1到len的数的二进制表示的1的位置，，。
太有意思了，。，哈哈

public class Solution {
    public void printAllCombination(String string) {
        if (string == null || string.length() == 0) {
            return;
        }
        char[] chars = string.toCharArray();
        int len = string.length();
        long end = (int) (Math.pow(2, len) - 1);
        for (int i = 1; i <= end; i++) {
            StringBuffer sb = new StringBuffer();
            String tempString = Integer.toBinaryString(i);
            while (tempString.length() < len) {
                tempString = "0" + tempString;
            }
            char[] tempChars = tempString.toCharArray();
            for (int j = 0; j < tempChars.length; j++) {
                if (tempChars[j] == '1') {
                    sb.append(chars[j]);
                }
            }
            System.out.println("result：" + sb.toString());
        }
    }
    public static void main(String[] args) {
        Solution solution = new Solution();
        String string = "abc";
        solution.printAllCombination(string);
    }
}