A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Find all strobogrammatic numbers that are of length = n.

For example,
Given n = 2, return ["11","69","88","96"].

Hint:

Try to use recursion and notice that it should recurse with n - 2 instead of n - 1.

246. Strobogrammatic Number 的變形，那道題讓我們判斷一個數是否是對稱數，而這道題讓我們找出長度為n的所有的對稱數。提示用遞歸來做，調用n-2，而不是n-1。

DFS，每次n-2，一層一層的叠代，到最後一層，n = 0 或者 n = 1 停止，backtracking加對稱數字。n = 0時加"‘, n = 1時，加0或1或8，其它層每次加能組成對稱數字對的一種組合，註意最外邊一層不能是0。

Java:

public class Solution {
    public List<String> findStrobogrammatic(int n) {
        return helper(n, n);
    }

    List<String> helper(int n, int m) {
        if (n == 0) return new ArrayList<String>(Arrays.asList(""));
        if (n == 1) return new ArrayList<String>(Arrays.asList("0", "1", "8"));

        List<String> list = helper(n - 2, m);
        List<String> res = new ArrayList<String>();

        for (int i = 0; i < list.size(); i++) {
            String s = list.get(i);
            if (n != m) res.add("0" + s + "0");
            res.add("1" + s + "1");
            res.add("6" + s + "9");
            res.add("8" + s + "8");
            res.add("9" + s + "6");
        }

        return res;
    }
}

Python:

class Solution:
    lookup = {‘0‘:‘0‘, ‘1‘:‘1‘, ‘6‘:‘9‘, ‘8‘:‘8‘, ‘9‘:‘6‘}

    def findStrobogrammatic(self, n):
        return self.findStrobogrammaticRecu(n, n)

    def findStrobogrammaticRecu(self, n, k):
        if k == 0:
            return [‘‘]
        elif k == 1:
            return [‘0‘, ‘1‘, ‘8‘]
        
        result = []
        for num in self.findStrobogrammaticRecu(n, k - 2):
            for key, val in self.lookup.iteritems():
                if n != k or key != ‘0‘:
                    result.append(key + num + val)

        return result

C++:

class Solution {
public:
    vector<string> findStrobogrammatic(int n) {
        return find(n, n);
    }
    vector<string> find(int m, int n) {
        if (m == 0) return {""};
        if (m == 1) return {"0", "1", "8"};
        vector<string> t = find(m - 2, n), res;
        for (auto a : t) {
            if (m != n) res.push_back("0" + a + "0");
            res.push_back("1" + a + "1");
            res.push_back("6" + a + "9");
            res.push_back("8" + a + "8");
            res.push_back("9" + a + "6");
        }
        return res;
    }
};

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