[leetcode] 247. Strobogrammatic Number II 解题报告_247. strobogrammatic number ii java-优快云博客

本文链接：https://blog.youkuaiyun.com/qq508618087/article/details/50946300

本文介绍了一种使用递归深度优先搜索(DFS)的方法来找出所有长度为n的旋转对称数。通过逐层构建字符串并考虑特殊字符的对称性，确保了生成的所有数在旋转180度后仍保持不变。

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题目链接: https://leetcode.com/problems/strobogrammatic-number-ii/

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.

思路: 可以像是一层层的给字符串从里向外穿衣服一样DFS生成所有的解.

其中翻转之后和自身相等有0, 1, 8, 在n为奇数的情况下最里面的一个数可以为这三个数的任意一个. 再外边就一次给两端添加一个对称的字符. 如果是最外层的话需要注意不能是为0.

代码如下:

class Solution {
public:
    void DFS(int n, string str)
    {
        if(n==0) return result.push_back(str);
        if(n%2==1) for(auto val: same) DFS(n-1, val);
        if(n%2==1) return;
        for(int i = (n==2)?1:0; i < two.size(); i++)
            DFS(n-2, two[i].first + str + two[i].second);
    }
    
    vector<string> findStrobogrammatic(int n) {
        if(n <= 0) return {};
        DFS(n, "");
        return result;
    }
private:
    vector<string> result;
    vector<string> same{"0", "1", "8"};
    vector<pair<char,char>> two{{'0','0'},{'1','1'},{'6','9'},{'8','8'},{'9','6'}};
};