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 = ; i <= num.length() / ; i++) {
             char a = num.charAt(i);
             char b = num.charAt(num.length() -  - i);
             if (!isValid(a, b)) {
                 return false;
             }
         }
         return true;
     }
     private boolean isValid(char c, char b) {
         switch (c) {
             case '':
                 return b == '';
             case '':
                 return b == '';
             case '':
                 return b == '';
             case '':
                 return b == '';
                 case '':
                 return b == '';
             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:
     vector<string> findStrobogrammatic(int n) {
         return find(n, n);
     }
     vector<string> find(int m, int n) {
         if (m == ) return {""};
         if (m == ) return {"", "", ""};
         vector<string> t = find(m - , n), res;
         for (auto a : t) {
             if (m != n) res.push_back("" + a + "");
             res.push_back("" + a + "");
             res.push_back("" + a + "");
             res.push_back("" + a + "");
             res.push_back("" + a + "");
         }
         return res;
     }
 };