LeetCode 246/247/248 中心对称数

LeetCode 246 (Easy)：判断一个数是否为中心对称数

LeetCode 247 (Medium)：给出长度为 n 的所有中心对称数

LeetCode 248 (Hard)：求给定区间内有多少中心对称数

LeetCode 246 (Easy)：判断一个数是否为中心对称数

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.

Examples:

Input:  "69"  Output: true
Input:  "88"  Output: true
Input:  "962" Output: false

解答：

两个数字只有这5种情况满足中心对称：11，88， 00， 69，96。可以看做求回文数的一种特殊情况，用双指针。

参考代码：https://www.cnblogs.com/grandyang/p/5196960.html

LeetCode 247 (Medium)：给出长度为 n 的所有中心对称数

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"].

Hint: Try to use recursion and notice that it should recurse with n - 2 instead of n - 1.

解答：

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: 1111, 6119, 8118, 9116, 1691, 6699, 8698, 9696,

1881, 6889, 8888, 9886, 1961, 6969, 8968, 9966, 1001, 6009, 8008, 9006

可以看到，长度为 i 的数是在长度为 i-2 的所有数的两端加上 11，88，69，96.

有一点要注意：长度为n的数，除了上述规律外，还多4个数 (因为长度为n-2的数两端不能为0)

因此，在中间递归(构造长度为i的数, i<n)的过程中，都需要在数字两端加上00，11，88，69，96。

最后构造长度为n的数时，只在两端加上11，88，69，96。

参考代码：https://www.cnblogs.com/grandyang/p/5200919.html

LeetCode 248 (Hard)：求给定区间内有多少中心对称数

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

Write a function to count the total strobogrammatic numbers that exist in the range of low <= num <= high.

Example:

Input: low = "50", high = "100"
Output: 3
Explanation: 69, 88, and 96 are three strobogrammatic numbers.

Note: Because the range might be a large number, the low and high numbers are represented as string.

解答：

从 n=0 和 n=1 的中心对称数开始，在每个数两端加上00，11，88，69，96。

得到的数num一定也是中心对称数，所以只需要判断num是否在[low, high]内即可( string可用运算符直接比较大小)。

class Solution {
public:string low, high;int cnt;int strobogrammaticInRange(string low, string high) {this->low = low;this->high = high;cnt = 0;solve("");solve("0");solve("1");solve("8");return cnt;}void solve(string num) {if((num.size() > high.size()) ||                 // 递归边界：若num的长度 > high的长度(num.size()==high.size() && num > high))      // 或长度相等,但num比high大 (string可用>直接比较)return;// 若num在[low,high]内, 就计数if(!(num.size()>1 && num[0]=='0') &&             // num开头不是0((num.size() > low.size()) ||(num.size()==low.size() && num>=low))        // 且num>=low (string可用>直接比较))++cnt;solve("0" + num + "0");solve("1" + num + "1");solve("8" + num + "8");solve("6" + num + "9");solve("9" + num + "6");}
};

参考代码：https://www.cnblogs.com/grandyang/p/5203228.html