LeetCode OJ - Letter Combinations of a Phone Number

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Note:

Although the above answer is in lexicographical order, your answer could be in any order you want.

思路一：hash表，DFS。

map<char, string> hash;

DFS(s, start, item) {
if(start == s.size()) {
ret += item;
return;
}
char cur = s[start];
for(i = start; i < hash[cur].size(); i++) {
item += hash[cur][i];
DFS(s, start + 1, item);
item -= hash[cur][i];
}
}

下面是AC代码，里面的测试用例输入"" 、期望输出[""]，感觉没什么道理

class Solution {
map<char, string> hash;
vector<string> ret;
public:
vector<string> letterCombinations(string digits) {
int len = digits.size();
if(len == 0) {
ret.push_back("");
return ret;
}
hash['0'] = "";
hash['1'] = "";
hash['2'] = "abc";
hash['3'] = "def";
hash['4'] = "ghi";
hash['5'] = "jkl";
hash['6'] = "mno";
hash['7'] = "pqrs";
hash['8'] = "tuv";
hash['9'] = "wxyz";

string item = "";
DFS(digits, 0, item);
return ret;
}

void DFS(string &digits, int start, string item) {
if(start == digits.size()) {
ret.push_back(item);
return ;
}
char cur = digits[start];
for(int i = 0; i < hash[cur].size(); i++) {
item.push_back(hash[cur][i]);
DFS(digits, start + 1, item);
item.pop_back();
}
}
};

思路二：组合问题，采用迭代方式解决。维护结果集ret，对于输入digits中的第i位对应字符串X，ret必须要根据当前状况和X对每一个ret[j]增加，即ret[j] + X[k]。因此是三维迭代。注意，这里ret[j] 有一个变为X.size()个，增加了X.size() - 1个。

class Solution {
public:
vector<string> letterCombinations(string digits) {
const string letters[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

vector<string> ret(1, "");
for (int i = 0; i < digits.size(); ++i) {
for (int j = ret.size() - 1; j >= 0; --j) {
const string &s = letters[digits[i] - '2'];  //取出string
for (int k = s.size() - 1; k >= 0; --k) {
if (k)
ret.push_back(ret[j] + s[k]);
else
ret[j] += s[k];
}
}
}

return ret;
}
};

此处要注意，ret.size是不断改变的，不能用来做循环判断（for j = 0 to ret.size() ）