力扣17.电话号码的字母组合

题目： 传送门
题意：

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 “任意顺序” 返回。

代码：

#include
#include
#include
#include
using namespace std;
string mp[10] = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
class Solution {
public:
	vector<string> letterCombinations(string digits) {
		vector<string>ans;
		queue<string>p;
		for (int i = 0; i < digits.size(); i++) {
			int x = digits[i]-'0';
			int len = p.size();
			if (len == 0) {
				for (int k = 0; k < mp[x].size(); k++) {
					string str1;
					str1.append(1,mp[x][k]);
					p.push(str1);
				}
			}
			for (int j = 0; j < len; j++) {
				string str;
				str = p.front();
				p.pop();
				for (int k = 0; k < mp[x].size(); k++) {
					string str1 = str;
					str1.append(1 , mp[x][k]);
					p.push(str1);
				}
				
			}
		}
		while (p.size()) {
			ans.push_back(p.front());
			p.pop();
		}
		return ans;
	}
	
};

int main() {
	string str;
	while (1) {
		cin >> str;
		vector<string>p;
		Solution s;
		p = s.letterCombinations(str);
		for (int i = 0; i < p.size(); i++)
			cout << p[i] << " ";
		cout << endl;
	}
	
	return 0;
}

可以使用队列，根据队列的特性，然后完成对所有情况的列举。如果把一个字符插入字符串，使用append（）函数来完成对字符串的更新。
注重容器的使用。

使用vector容器：

#include
#include
#include
#include
using namespace std;
string mp[10] = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
class Solution {
public:
	vector<string> letterCombinations(string digits) {
		vector<string>ans;
		for (int i = 0; i < digits.size(); i++) {
			int x = digits[i] - '0';
			int len = ans.size();
			if (len == 0) {
				for (int k = 0; k < mp[x].size(); k++) {
					string str;
					str.append(1, mp[x][k]);
					ans.push_back(str);
				}
			}
			else {
				for (int i = 0; i < len; i++) {
					string str = ans[0];
					for (int k = 0; k < mp[x].size(); k++) {
						string str1=str;
						str1.append(1, mp[x][k]);
						ans.push_back(str1);
					}
					vector< string >::iterator k = ans.begin();///删除第一个元素
					ans.erase(k);///删除指定元素
				}
			}
		}
		return ans;
	}
	
};

int main() {
	string str;
	while (1) {
		cin >> str;
		vector<string>p;
		Solution s;
		p = s.letterCombinations(str);
		for (int i = 0; i < p.size(); i++)
			cout << p[i] << " ";
		cout << endl;
	}
	
	return 0;
}

使用dfs深搜：

#include
#include
#include
#include
using namespace std;
string mp[10] = { "","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz" };
class Solution {
public:
	vector<string>ans;
	string str;
	void dfs(int x,string digits) {
		if (x == digits.size()) {
			ans.push_back(str);
			return;
		}
		int k = digits[x] - '0';
		for (int i = 0; i < mp[k].size(); i++) {
			string str1 = str;
			str.append(1, mp[k][i]);
			dfs(x+1, digits);
			str = str1;
		}
		return;
	}
	vector<string> letterCombinations(string digits) {
		if (digits.size() == 0)
			return ans;
		dfs(0, digits);
		return ans;
	}
	
};

int main() {
	string str;
	while (1) {
		cin >> str;
		vector<string>p;
		Solution s;
		p = s.letterCombinations(str);
		for (int i = 0; i < p.size(); i++)
			cout << p[i] << " ";
		cout << endl;
	}
	
	return 0;
}