题目描述：
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.
Example:
Input: n = 2
Output: [“11”,“69”,“88”,“96”]
Time complexity: O(n)
观察规律：
n == 1: 0, 1, 8
n == 2: 11, 88, 69, 96
n ==3：
101 111 181
808 818 888
609 619 689
906 916 986
n ==4：
1111 1881 1691 1961 1001
8118 8888 8698 8968 8008
6119 6889 6699 6969 6009
9116 9886 9696 9966 9006
对于 n>= 3 来说：如果 字符串s 是一个 strobogrammatic 则
1s1, 8s8, 6s9, 9s6, 0s0 都是strobogrammatic。 所以我们可以递归每次从最前端和最后端缩短两个字符。注意0不能作为开头。
class Solution {
private static final char[][] PAIRS = new char[][] {
{
'0', '0'}, {
'1', '1'}, {
'6', '9'}, {
'8', '8'}, {
'9', '6'}};
List<String> res = new ArrayList<>();
public List<String> findStrobogrammatic(int n) {
if (n == 0) return Arrays.asList("");
if (n == 1) return Arrays.asList("0", "1", "8");
dfs(new char[n], 0, n-1);
return res;
}
void dfs(char[] buffer, int left, int right){
if (left > right) {
res.add(String.valueOf(buffer));
return;
}
for(char[] p : PAIRS){
if (buffer.length != 1 && left == 0 && p[0] == '0') {
continue;
}
if (left == right && (p[0] == '6' || p[0] == '9')) {
continue;
}
buffer[left] = p[0];
buffer[right] = p[1];
dfs(buffer, left+1, right-1);
}
}
}
class Solution {
public List<String> findStrobogrammatic(int n) {
return helper(n, n);
}
List<String> helper(int n, int len){
if (n == 0) return Arrays.asList("");
if (n == 1) return Arrays.asList("0", "1", "8");
List<String> list = helper(n - 2, len);
List<String> res = new ArrayList<String>();
for(String s:list){
if (n != len) {
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;
}
}