247.Strobogrammatic Number II

    /*
     * 247.Strobogrammatic Number II
     * 2016-6-18 by Mingyang
     * 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"].
     * 利用了backtracking的概念，其实就是dfs
     * 非常好的一道backtracking的题目，因为这里需要区分我们的奇数偶数个，所以我们需要分别来讨论
     * 关键点1:
     * for(char c : map.keySet())
     * 关键点2：
     * sb.insert(0, c);
     * sb.append(map.get(c));
     * sb.deleteCharAt(0);
     * sb.deleteCharAt(sb.length() - 1); 
     */
     public List<String> findStrobogrammatic1(int n) {
            if(n < 1)  return new ArrayList<String>();
            List<String> res = new ArrayList<String>();
            Map<Character, Character> map = new HashMap<Character, Character>();
            map.put(‘0‘, ‘0‘);
            map.put(‘1‘, ‘1‘);
            map.put(‘6‘, ‘9‘);
            map.put(‘8‘, ‘8‘);
            map.put(‘9‘, ‘6‘);       
            StringBuilder sb = new StringBuilder();
            int position = (n % 2 == 0) ? 0 : 1;
            dfs(res, sb, map, n, position);        
            return res;
        }    
        private void dfs(List<String> res, StringBuilder sb, Map<Character, Character> map, int n, int position) {
            if(sb.length() > n)
                return;
            if(sb.length() == n) {
                res.add(sb.toString());
                return;
            }      
            if(position == 1) {
                for(char c : map.keySet()) {
                    if(c == ‘6‘ || c == ‘9‘)
                        continue;
                    sb.append(c);
                    dfs(res, sb, map, n, position + 1);
                    sb.setLength(0);
                }
            } else {
                for(char c : map.keySet()) {
                    if(n - sb.length() == 2 && c == ‘0‘)
                        continue;
                    sb.insert(0, c);
                    sb.append(map.get(c));
                    dfs(res, sb, map, n, position + 2);
                    sb.deleteCharAt(0);
                    sb.deleteCharAt(sb.length() - 1);
                }    
            }
        }
     //网上的改进型代码：
      public List<String> findStrobogrammatic(int n) {
            Map<Character, Character> map = new HashMap<Character, Character>();
            map.put(‘0‘, ‘0‘);
            map.put(‘1‘, ‘1‘);
            map.put(‘6‘, ‘9‘);
            map.put(‘8‘, ‘8‘);
            map.put(‘9‘, ‘6‘);
            List<String> result = new ArrayList<String>();
            char[] buffer = new char[n];
            dfs(n, 0, buffer, result, map);
            return result;
       }
       private void dfs(int n, int index, char[] buffer, List<String> result, Map<Character, Character> map) {
            if (n == 0) {
                return;
            }
            if (index == (n + 1) / 2) {
                result.add(String.valueOf(buffer));
                return;
            }
            for (Character c : map.keySet()) {
                if (index == 0 && n > 1 && c == ‘0‘) {  // first digit cannot be ‘0‘ when n > 1
                    continue;
                }
                if (index == n / 2 && (c == ‘6‘ || c == ‘9‘)) {   // mid digit cannot be ‘6‘ or ‘9‘ when n is odd
                    continue;
                }
                buffer[index] = c;
                buffer[n - 1 - index] = map.get(c);
                dfs(n, index + 1, buffer, result, map);
            }
       }