[数据结构与算法]剑指offer--38字符串的排列

题目

代码

class Solution {
    public String[] permutation(String s) {
        List<String> list = new ArrayList();
        char[] arr = s.toCharArray();
        StringBuilder sb = new StringBuilder();
        boolean[] visited = new boolean[arr.length];
        dfs(arr, "", visited, list);
        //new String[0]只能填0，代表类型的规定，其他的数就限定了输出的维度，java8如此
        return list.toArray(new String[0]);
    }
    public void dfs(char[] arr, String s,  boolean[] visited, List list)
    {
        if(s.length() == arr.length)
        {
            list.add(s);
            return;
        }
        for(int i=0; i<arr.length; i++)
        {
            if(visited[i]) continue;
            visited[i] = true;
            dfs(arr, s+String.valueOf(arr[i]), visited, list);
            visited[i] = false;  //回溯
        }

    }
}

结果