leetcode 247

问题

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.

Example:

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Input:  n = 2
Output: ["11","69","88","96"]

分析

思路1：先考虑n位数的前n/2位，然后在通过前n/2位生成后n/2。注意corner case，n为奇数的情况下这n位数的最中间的一位只能取到’0’, ‘1’, ‘8’三种情况，以及在考虑前n/2时，第一位应该从’1’开始取，其他位从’0’开始取。

思路2：采用确定中间进行两边扩展的方式，递归地进行处理。注意corner case，n为奇数和n为偶数，以及要不要首位添加’0’的判断。

代码1

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class Solution {
public:
    char a[5] = {'0', '1', '6', '8', '9'};
    vector<string> findStrobogrammatic(int n) {
        vector<string> ans;
        int suffixNum = n/2;
        bool odd = (n%2)==1? true: false;
        string s;
        dfs(ans, s, suffixNum, odd);
        return ans;
    }
    
    void dfs(vector<string> &ans, string &s, int suffixNum, bool odd) {
        if (s.size() == suffixNum) {
            string rs = getMirror(s);
            if (odd) {
                ans.push_back(s + "0" + rs);
                ans.push_back(s + "1" + rs);
                ans.push_back(s + "8" + rs);
            } else {
                ans.push_back(s+rs);
            }
        } else {
            int startIndex = s.size() == 0? 1: 0;
            for (int i = startIndex; i < 5; ++i) {
                s += a[i];
                dfs(ans, s, suffixNum, odd);
                s.pop_back();
            }
        }
    }

    string getMirror(const string &s) {
        string ret;
        for (int i = s.size()-1; i>=0; --i) {
            char c;
            switch (s[i]) {
                case '0': c = '0'; break;
                case '1': c = '1'; break;
                case '6': c = '9'; break;
                case '8': c = '8'; break;
                case '9': c = '6'; break;
                default: break;
            }
            ret += c;
        }
        return ret;
    }
};

代码2

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class Solution {
public:
    vector<string> findStrobogrammatic(int n) {
        return helper(n, n);
    }
    vector<string> helper(int n, int m) {
        if (n == 0) return {""};
        if (n == 1) return {"0", "1", "8"};
        
        vector<string> number = helper(n-2, m);
        vector<string> ret;
        for (string &s: number) {
            if (n != m)
                ret.push_back('0' + s + '0');
            ret.push_back('1'+s+'1');
            ret.push_back('6'+s+'9');
            ret.push_back('8'+s+'8');
            ret.push_back('9'+s+'6');
        }
        return ret;
    }
};