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 的变形，那道题让我们判断一个数是否是对称数，而这道题让我们找出长度为n的所有的对称数。提示用递归来做，调用n-2，而不是n-1。
DFS，每次n-2，一层一层的迭代，到最后一层，n = 0 或者 n = 1 停止，backtracking加对称数字。n = 0时加"‘, n = 1时，加0或1或8，其它层每次加能组成对称数字对的一种组合，注意最外边一层不能是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;
}
};
类似题目：
[LeetCode] 246. Strobogrammatic Number 对称数
[LeetCode] 248. Strobogrammatic Number III 对称数III
