剑指 Offer 38. 字符串的排列

public String[] permutation(String s) {
        //显然是递归加回溯
        char[] c = s.toCharArray();
        int n = c.length;
        //可能有重复元素，所以用set来存储，
        Set<String> set = new HashSet<>();
        char[] list = new char[n];
        boolean[] visited = new boolean[n];
        dfs(c,list,visited,0,set,n);
        String[] res = new String[set.size()];
        int index = 0;
        Iterator<String> iterator = set.iterator();
        while (iterator.hasNext()){
            String st = iterator.next();
            res[index++] = st;
            iterator.remove();
        }
        return res;
    }

    private void dfs(char[] c, char[] list, boolean[] visited, int count, Set<String> set,int n) {
        if(count==n){
            String s = new String(list);
            set.add(s);
        }
        for(int i = 0;i<n;i++){
            if(!visited[i]){
                list[count] = c[i];
                visited[i] = true;
                dfs(c,list,visited,count+1,set,n);
                visited[i] = false;
            }
        }
    }