LeetCode-Strobogrammatic Number II
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"]
.
Solution:
1 public class Solution { 2 static char[] nums1 = new char[] { '0', '1', '8', '6', '9' }; 3 static char[] nums2 = new char[] { '0', '1', '8', '9', '6' }; 4 5 public List<String> findStrobogrammatic(int n) { 6 List<String> res = new ArrayList<String>(); 7 StringBuilder builder = new StringBuilder(); 8 findStrobogrammaticRecur(n, builder, 1, res); 9 10 return res; 11 } 12 13 public void findStrobogrammaticRecur(int n, StringBuilder builder, int curLevel, List<String> res) { 14 if (curLevel > n / 2) { 15 if (n % 2 == 1) { 16 // add one more char into front, no '6' or '9' 17 for (int i = 0; i < 3; i++) { 18 builder.insert(n / 2, nums1[i]); 19 res.add(builder.toString()); 20 builder.deleteCharAt(n / 2); 21 } 22 } else 23 res.add(builder.toString()); 24 return; 25 } 26 27 // if curLevel is n/2, we should skip '0'. 28 int start = (curLevel == n/2) ? 1 : 0; 29 for (int i = start; i < 5; i++) { 30 builder.insert(0, nums1[i]); 31 builder.append(nums2[i]); 32 findStrobogrammaticRecur(n, builder, curLevel + 1, res); 33 builder.deleteCharAt(0); 34 builder.deleteCharAt(builder.length() - 1); 35 } 36 } 37 }