回溯法(算法)

 

leecode第17题

从上至下遍历每个节点，直至最深深度为方法出口，返回上一层，然后移除当前元素。

 

 

public class test {
    public static void main(String[] args) {
        String s = "23";
        List<String> list = new test().solution(s);
        System.out.println(list);
    }

    private List<String> solution(String digits) {
        List<String> combinations = new ArrayList<>();
        if (digits.length() == 0) {
            return combinations;
        }

        Map<Character, String> phoneMap = new HashMap<Character, String>() {{
            put('2', "abc");
            put('3', "def");
            put('4', "ghi");
            put('5', "jkl");
            put('6', "mno");
            put('7', "pqrs");
            put('8', "tuv");
            put('9', "wxyz");
        }};

        int p = 0;
        StringBuilder sb = new StringBuilder();
        List result = exectorSolution(phoneMap, digits, combinations, p, sb);
        return result;


    }

    private List exectorSolution(Map<Character, String> phoneMap, String digits, List<String> combinations, int p, StringBuilder sb) {
        //若索引遍历到底后，递归方法结束
        if (digits.length() == p) {
            //达到最深，list中添加sb其中一种组合情况
            combinations.add(sb.toString());
        } else {
            String str = phoneMap.get(digits.charAt(p));
            //遍历当前深度对应的所有字符
            for (int i = 0; i < str.length(); i++) {
                sb.append(str.charAt(i));
                //回溯法，调用下一个节点深度的所有字符
                exectorSolution(phoneMap, digits, combinations, p + 1, sb);
                //结束后把当前的字符串从sb中移除
                sb.deleteCharAt(p);
            }

        }
        return combinations;
    }
}