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"], A solution is:
[ ["abc","bcd","xyz"], ["az","ba"], ["acef"], ["a","z"] ] 字符串的每个字母和首字符的相对距离都是相等的，比如abc和efg互为偏移，对于abc来说，b和a的距离是1，c和a的距离是2，对于efg来说，f和e的距离是1，g和e的距离是2。再来看一个例子，az和yx，z和a的距离是25，x和y的距离也是25(直接相减是-1，这就是要加26然后取余的原因)，那么这样的话，所有互为偏移的字符串都有个unique的距离差，我们根据这个来建立映射就可以很好的进行单词分组了
class Solution {
public:
vector<vector<string>> groupStrings(vector<string>& strings) {
vector<vector<string> > res;
unordered_map<string, multiset<string>> m;
for (auto a : strings) {
string t = "";
for (char c : a) {
t += to_string((c + 26 - a[0]) % 26) + ",";
}
m[t].insert(a);
}
for (auto it = m.begin(); it != m.end(); ++it) {
res.push_back(vector<string>(it->second.begin(), it->second.end()));
}
return res;
}
};