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：
1 class Solution {
2 public:
3 vector<string> findStrobogrammatic(int n) {
4 return find(n, n);
5 }
6 vector<string> find(int m, int n) {
7 if (m == 0) return {""};
8 if (m == 1) return {"0", "1", "8"};
9 vector<string> 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 };