My code:

public class Solution {
    public boolean isStrobogrammatic(String num) {
        if (num == null || num.length() == 0) {
            return true;
        }
        
        HashMap<Character, Character> map = new HashMap<Character, Character>();
        map.put('6', '9');
        map.put('9', '6');
        map.put('1', '1');
        map.put('0', '0');
        map.put('8', '8');
        
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < num.length(); i++) {
            char curr = num.charAt(i);
            if (!map.containsKey(curr)) {
                return false;
            }
            else {
                sb.append(map.get(curr));
            }
        }
        
        return sb.reverse().toString().equals(num);
    }
}

比较简单的题。
可以再优化。hashmap改用array，循环改用双指针。

My code:

public class Solution {
    public boolean isStrobogrammatic(String num) {
        if (num == null || num.length() == 0) {
            return true;
        }
        
        int[] map = new int[10];
        Arrays.fill(map, -1);
        map[0] = 0;
        map[1] = 1;
        map[6] = 9;
        map[8] = 8;
        map[9] = 6;
        int begin = 0;
        int end = num.length() - 1;
        while (begin <= end) {
            char c1 = num.charAt(begin);
            char c2 = num.charAt(end);
            if (map[c1 - '0'] != c2 - '0') {
                return false;
            }
            begin++;
            end--;
        }
        return true;
    }
}

reference:
https://discuss.leetcode.com/topic/55075/0-ms-array-based-java-solution-easy-to-understand

Anyway, Good luck, Richardo! — 09/22/2016