[leetcode 249] Group Shifted Strings

Question:

Given a string, we can "shift" each of its letter to its successive letter, for example:"abc" -> "bcd". We can keep "shifting" which forms the sequence:

"abc" -> "bcd" -> ... -> "xyz"

Given a list of strings which contains only lowercase alphabets, group all strings that belong to the same shifting sequence.

For example, given:["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"],

Return:

[

["abc","bcd","xyz"],

["az","ba"],

["acef"],

["a","z"]

]

分析：

使用hashmap，关键字key为首个字符串相邻元素的差值和，如果差值为负数，则加上26，比如“abc”的key是（‘a’-‘b’+26）+（‘b’-‘c’+26），再转化成字符串。

这样，shifted字符串的关键字key相同。

代码如下：

<span style="font-size:14px;">class Solution {
public:
vector<vector<string>> groupStrings(vector<string>& strings) {
unordered_map<string, vector<string>> hashmap;
vector<vector<string>> res;

for (int i = 0; i < strings.size(); ++i) {
hashmap[calKey(strings[i])].push_back(strings[i]);
}

for (auto it = hashmap.begin(); it != hashmap.end(); ++it) {
sort(it->second.begin(), it->second.end());//second 访问的是map的第二个元素，即元素值。
vector<string> tmp = it->second;
res.push_back(tmp);
}

return res;
}
private:
string calKey(string input) {
string key = "";
for (int i = 0; i < input.length() - 1; ++i) {
int gap = (input[i] > input[i+1]) ? (input[i] - input[i+1]) : (input[i] - input[i+1] + 26);
char num = 'a' + gap;
key += num;
}
return key;
}
};</span>