247. Strobogrammatic Number

Strobogrammatic Number I

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Write a function to determine if a number is strobogrammatic. The number is represented as a string.

For example, the numbers "69", "88", and "818" are all strobogrammatic.

From:https://segmentfault.com/a/1190000003787462

翻转后对称的数就那么几个，我们可以根据这个建立一个映射关系：8->8, 0->0, 1->1, 6->9, 9->6，然后从两边向中间检查对应位置的两个字母是否有映射关系就行了。比如619，先判断6和9是有映射的，然后1和自己又是映射，所以是对称数。

public class Solution {
    public boolean isStrobogrammatic(String num) {
        for (int i = 0; i <= num.length() / 2; i++) {
            char a = num.charAt(i);
            char b = num.charAt(num.length() - 1 - i);
            if (!isValid(a, b)) {
                return false;
            }
        }
        return true;
    }
    private boolean isValid(char c, char b) {
        switch (c) {
            case '1':
                return b == '1';
            case '6':
                return b == '9';
            case '9':
                return b == '6';
            case '8':
                return b == '8';
                case '0':
                return b == '0';
            default:
                return false;
        }
    }
}

Strobogrammatic Number II

From: http://www.cnblogs.com/grandyang/p/5200919.html

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"].

这道题是之前那道Strobogrammatic Number的拓展，那道题让我们判断一个数是否是对称数，而这道题让我们找出长度为n的所有的对称数，我们肯定不能一个数一个数的来判断，那样太不高效了，而且OJ肯定也不会答应。题目中给了提示说可以用递归来做，而且提示了递归调用n-2，而不是n-1。我们先来列举一下n为0,1,2,3,4的情况：

n = 0:   none

n = 1:   0, 1, 8

n = 2:   11, 69, 88, 96

n = 3:   101, 609, 808, 906, 111, 619, 818, 916, 181, 689, 888, 986

n = 4:   1001, 6009, 8008, 9006, 1111, 6119, 8118, 9116, 1691, 6699, 8698, 9696, 1881, 6889, 8888, 9886, 1961, 6969, 8968, 9966

我们注意观察n=0和n=2，可以发现后者是在前者的基础上，每个数字的左右增加了[1 1], [6 9], [8 8], [9 6]，看n=1和n=3更加明显，在0的左右增加[1 1]，变成了101, 在0的左右增加[6 9]，变成了609, 在0的左右增加[8 8]，变成了808, 在0的左右增加[9 6]，变成了906, 然后在分别在1和8的左右两边加那四组数，我们实际上是从m=0层开始，一层一层往上加的，需要注意的是当加到了n层的时候，左右两边不能加[0 0]，因为0不能出现在两位数及多位数的开头，在中间递归的过程中，需要有在数字左右两边各加上0的那种情况，参见代码如下：  

class Solution {
    public List<String> findStrobogrammatic(int n) {
        return helper(n, n);
    }

    List<String> helper(int n, int m) {
        if (n == 0) return new ArrayList<>(Arrays.asList(""));
        if (n == 1) return new ArrayList<>(Arrays.asList("0", "1", "8"));
      // we added 2 more digits,that is why n-2
        List<String> list = helper(n - 2, m);
        List<String> res = new ArrayList<>();

        for (int i = 0; i < list.size(); i++) {
            String s = list.get(i);
            // 0 cannot be placed at the beginning and ending of the digits
            if (n != m) 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;
    }
}