这道题要经常拿出来重新看一看,动手编一编!
和leetcode第39题组合总和的思路类似。
思路:回溯法
#include<iostream>
#include<map>
#include<vector>
#include<iterator>
#include<string>
using namespace std;
map<char, string> mp = { { '2',"abc" },{ '3',"def" },{ '4',"ghi" },{ '5',"jkl" },{ '6',"mno" },{ '7',"pqrs" },{ '8',"tuv" },{ '9',"wxyz" } };
class Solution {
public:
vector<string> letterCombinations(string numberstr) {
if (numberstr.size() == 0)
return res;
DFS(numberstr);
return res;
}
void DFS(string numberstr)
{
if (numberstr.size() == 0)//递归结束的条件
{
res.push_back(cur);
return;
}
string letterstr = mp[numberstr[0]];
for (int i = 0; i < letterstr.size(); i++) {
cur.push_back(letterstr[i]);
DFS(numberstr.substr(1));//substr函数返回的是对象,不是引用
cur.pop_back();//核心:上一句总是会出来的
}
}
private:
vector<string> res;
string cur;
};
int main() {
Solution s;
vector<string>result = s.letterCombinations("23");
copy(result.begin(), result.end(), ostream_iterator<string>(cout, " "));
system("pause");
return 0;
}