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); 
     */

     List<String> res=new ArrayList<String>();
        public  List<String> findStrobogrammatic(int n) {
              if(n<=0)
              return res;
          HashMap<Character, Character> map = new HashMap<Character, Character>();
                map.put('1','1');
                map.put('0','0');
                map.put('6','9');
                map.put('9','6');
                map.put('8','8');
            int position = (n % 2 == 0) ? 0 : 1;
            dfs(n,new StringBuffer(),map,position);
            return res;
        }
        public  void dfs(int n,StringBuffer sb,HashMap<Character,Character> map,int position){
            if(n<0)
                return;
            if(n==0){
                res.add(sb.toString());
                return;
            }
             if(position == 1) {
                    for(char c : map.keySet()) {
                        if(c == '6' || c == '9')
                            continue;
                        sb.append(c);
                        dfs(n-1, sb, map,position + 1);
                        sb.setLength(0);
                    }
             }else{
                 for (Map.Entry<Character, Character> entry : map.entrySet()) {
                     if(n  == 2 && entry.getKey() == '0')
                            continue;
                 sb.insert(0,entry.getKey());
                 sb.append(entry.getValue());
                 dfs(n-2,sb,map,position);
                 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);
            }
       }