869. 重新排序得到 2 的幂

链接：869. 重新排序得到 2 的幂

题解：

class Solution {
public:
    bool reorderedPowerOf2(int n) {
        std::string str = std::to_string(n);
        sort(str.begin(), str.end());
        int begin = 0;
        std::vector visited(str.size(), false);
        std::vector path;
        int sum = 0;
        return dfs(begin, str, path, visited, sum);
        
    }
    bool dfs(int begin, const std::string& str, std::vector& path, std::vector& visited, int sum) {
        if (begin >= str.size()) {
            return two(sum);
        }
        for (int i = 0; i < str.size(); ++i) {
            //cout << str[i] << endl;
            if (begin == 0 && str[i] == '0') {
                continue;
            }
            if (visited[i]) {
                continue;
            }
            if (i-1 >= 0 && !visited[i-1] && str[i] == str[i-1]) {
                continue;
            }
            
            visited[i] = true;
            path.push_back(str[i]);
            if (dfs(begin+1, str, path, visited, sum*10 + (str[i]-'0'))) {
                return true;
            }
            path.pop_back();
            visited[i] = false;
        }
        return false;
    }
private:
    bool two(int n) {
        if (n == 1) {
            return true;
        }
        return (n&(n-1)) == 0 ? true : false;
    }
};