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。

1 public class Solution { 2     public boolean isStrobogrammatic(String num) { 3         for (int i = 0; i <= num.length() / 2; i++) { 4             char a = num.charAt(i); 5             char b = num.charAt(num.length() - 1 - i); 6             if (!isValid(a, b)) { 7                 return false; 8             } 9         }10         return true;11     }12     private boolean isValid(char c, char b) {13         switch (c) {14             case '1':15                 return b == '1';16             case '6':17                 return b == '9';18             case '9':19                 return b == '6';20             case '8':21                 return b == '8';22                 case '0':23                 return b == '0';24             default:25                 return false;26         }27     }28 }

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：  

1 class Solution { 2 public: 3     vector
     
       findStrobogrammatic(int n) { 4         return find(n, n); 5     } 6     vector
      
        find(int m, int n) { 7         if (m == 0) return {
   ""}; 8         if (m == 1) return {
   "0", "1", "8"}; 9         vector
       
         t = find(m - 2, n), res;10         for (auto a : t) {11             if (m != n) res.push_back("0" + a + "0");12             res.push_back("1" + a + "1");13             res.push_back("6" + a + "9");14             res.push_back("8" + a + "8");15             res.push_back("9" + a + "6");16         }17         return res;18     }19 };