LeetCode 题解(239) : Strobogrammatic Number II

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题目：

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 withn - 2 instead of n - 1.

题解：

C++版和Java版是递归，Python版是循环。

C++版：

class Solution {public:    vector<string> findStrobogrammatic(int n) {        vector<string> results;        if(n == 0)            return results;                    unordered_map<char, string> d;        d.insert(pair<char, string>('0', "0"));        d.insert(pair<char, string>('1', "1"));        d.insert(pair<char, string>('6', "9"));        d.insert(pair<char, string>('8', "8"));        d.insert(pair<char, string>('9', "6"));                char s[] = {'0', '1', '8', '6', '9'};        int mid = n / 2;        string result = "";        bool oneMore = (bool)(n % 2);        recursion(d, results, result, mid, 0, s, oneMore);        return results;    }        void recursion(unordered_map<char, string> &d, vector<string> &results, string result, int recur, int iter, char* s, bool &oneMore) {        if(iter > recur) {            for(int i = 0; i < 3; i++) {                results.push_back(result.substr(0, recur) + s[i] + result.substr(recur));            }            return;        }        if(iter == recur) {            if(oneMore) {                recursion(d, results, result, recur, iter + 1, s, oneMore);            } else {                results.push_back(result);            }            return;        }        if(iter == 0) {            for(int i = 1; i < 5; i++) {                recursion(d, results, s[i] + d[s[i]], recur, iter + 1, s, oneMore);            }        } else {            for(int i = 0; i < 5; i++) {                recursion(d, results, result.substr(0, iter) + s[i] + d[s[i]] + result.substr(iter), recur, iter + 1, s, oneMore);            }        }    }};

Java版：

public class Solution {    public List<String> findStrobogrammatic(int n) {        List<String> results = new ArrayList<>();        if(n == 0)            return results;                    String result = "";        HashMap<Character, String> d = new HashMap<>();        d.put('0', "0");        d.put('1', "1");        d.put('6', "9");        d.put('8', "8");        d.put('9', "6");        String s = "01869";        int mid = n / 2;        int oneMore = n % 2;        recursion(d, results, result, s, mid, 0, oneMore);        return results;    }        public void recursion(HashMap<Character, String> d, List<String> results, String result, String s, int recur, int iter, int oneMore) {        if(iter > recur) {            for(int i = 0; i < 3; i++) {                results.add(result.substring(0, recur) + s.charAt(i) + result.substring(recur));            }            return;        }        if(iter == recur) {            if(oneMore == 1) {                recursion(d, results, result, s, recur, iter + 1, oneMore);            } else {                String local = result;                results.add(local);            }            return;        }        if(iter == 0) {            for(int i = 1; i < 5; i++) {                recursion(d, results, s.charAt(i) + d.get(s.charAt(i)), s, recur, iter + 1, oneMore);            }        } else {            for(int i = 0; i < 5; i++) {                recursion(d, results, result.substring(0, iter) + s.charAt(i) + d.get(s.charAt(i)) + result.substring(iter), s, recur, iter + 1, oneMore);            }        }    }}

Python版：

from copy import copyclass Solution(object):    def findStrobogrammatic(self, n):        """        :type n: int        :rtype: List[str]        """        s = ['0', '1', '8', '6', '9']        d = {'0': '0', '1': '1', '6': '9', '8': '8', '9': '6'}        ret1, ret2 = [""], []        if n == 0:            return ret1                    mid = n / 2        i = 0        while i < mid:            if i == 0:                for j in range(1, len(s)):                    ret2.append(s[j] + d[s[j]])            else:                for j in ret1:                    for k in s:                        ret2.append(j[:i] + k + d[k] + j[i:])            ret1 = copy(ret2)            ret2 = []            i += 1                    if n % 2 != 0:            for j in ret1:                for k in range(3):                    ret2.append(j[:i] + s[k] + j[i:])            ret1 = copy(ret2)                            return ret1

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