自学算法之字符的组合

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字符的组合

  • 题目:
    1. 输入一个字符串,打印出该字符串中字符的所有排列。例如输入字符串abc,则输出由字符a、b、c所能排列出来的所有字符串abc、acb、bac、bca、cab和cba。 题目:
  • 方法一:暴力遍历
        public static void main(String[] args) {
            Scanner in = new Scanner(System.in);
            while(in.hasNext()) {
                String inStr = in.next();
                ArrayList<String> result = new ArrayList<>();
                // SubString返回 i , j - 1的字符
                for(int i = 0; i < inStr.length(); i++) {
                    for(int j = i + 1; j <= inStr.length(); j++) {
                        String temp = inStr.substring(i, j);
                        if(!result.contains(temp)) {
                            result.add(temp);
                        }
                    }
                }
                Collections.sort(result, (a, b) -> a.length() != b.length() ? a.length() - b.length() : a.compareTo(b));
                for(String val : result) {
                    System.out.println(val);
                }
            }
        }
  • 方法二:递归求解
        @Test
        public void test() {
            Scanner in = new Scanner(System.in);
            while(in.hasNext()) {
                String inStr = in.next();
                ArrayList<String> result = new ArrayList<>();
                Method(result, new StringBuilder(), 0, inStr);
                Collections.sort(result, (a, b) -> a.length() != b.length() ? a.length() - b.length() : a.compareTo(b));
                for(String val : result) {
                    System.out.println(val);
                }
            }
        }
        public void Method(ArrayList<String> result, StringBuilder sb, int start, String str) {
            // 集合中包含重复值,递归结束
            if(result.contains(sb.toString())) {
                return ;
            }
            // 一个字符, 直接入集合reuslt
            if(sb.length() == 1) {
                result.add(sb.toString());
            } else if(sb.length() > 1 && str.contains(sb.toString())) {
                // 
                result.add(sb.toString());
            }
            for(int i = start; i < str.length(); i++) {
                sb.append(str.charAt(i));
                Method(result, sb, i + 1, str);
                sb.deleteCharAt(sb.length() - 1);
            }
        }

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转载自blog.csdn.net/qq_37014990/article/details/82708464