刷题17. Letter Combinations of a Phone Number

一、题目说明

题目17. Letter Combinations of a Phone Number，题目给了下面一个图，输入一个字符串包括2-9，输出所有可能的字符组合。

如输入23所有可能的输出：

"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"

二、我的做法

这个题目，我思考了4个小时（惭愧严重超时了），做法如下：

#include<iostream>
#include<vector>
#include<unordered_map>
using namespace std;
class Solution{
	public:
		vector<string> letterCombinations(string s){
			vector<string> res;

			if(s.size()<1) return res;
			int num = 1;
			unordered_map<char,string> ump;
			
			ump['2'] = "abc";
			ump['3'] = "def";
			ump['4'] = "ghi";
			ump['5'] = "jkl";
			ump['6'] = "mno";
			ump['7'] = "pqrs";
			ump['8'] = "tuv";
			ump['9'] = "wxyz";
			
			for(int i=0;i<s.size();i++){
				switch(s[i]){
					case '2':
					case '3':
					case '4':
					case '5':
					case '6':
					case '8':
						num *= 3;
						break;
					case '7':
					case '9':
					    num *=4;
						break;					
				}
			}
			for(int i=0;i<num;i++){
				res.push_back("");
			}

            int curNum = num;
			for(int j=0;j<s.size();++j){
				char curr = s[j];
				string curStr = ump[curr];
				curNum /= curStr.size();
			    for(int i=0;i<num;i++){
			    	res[i].push_back(curStr[i / curNum % curStr.size()]);
				}
			}

			return res;
		} 
};
int main(){
	Solution s;
//    vector<string> r = s.letterCombinations("234");
//    for(vector<string>::iterator it=r.begin();it!=r.end();++it){
//		cout<<*it<<" ";
//	}
//	cout<<endl;
	vector<string> r = s.letterCombinations("8");
    for(vector<string>::iterator it=r.begin();it!=r.end();++it){
		cout<<*it<<" ";
	}
	return 0;
}

这个是我第一次，做“完美”的代码。臭美一下！

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Letter Combinations of a Phone Number.
Memory Usage: 8.4 MB, less than 100.00% of C++ online submissions for Letter Combinations of a Phone Number.

三、更优化的做法

第一次可以自豪的说一句，这个就是最优化的代码了。哈哈！