[leetcode][hash] Anagrams

题目;

Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.
v1 76ms

class Solution {
public:
//将单词按字母顺序排序后作为hash的key，字符串在strs中的下标组成的vector作为value
vector<string> anagrams(vector<string>& strs) {
vector<string> res;
map<string, vector<int> > hashTable;
for (int i = 0; i < strs.size(); ++i){
string s = strs[i];
sort(s.begin(), s.end());
map<string, vector<int> >::iterator iter = hashTable.find(s);
if (iter != hashTable.end()){
(iter->second).push_back(i);
}
else{
vector<int> vTmp;
vTmp.push_back(i);
hashTable[s] = vTmp;
}
}
for (map<string, vector<int> >::iterator iter = hashTable.begin(); iter != hashTable.end(); ++iter){
if ((iter->second).size() > 1){
for (int i = 0; i < (iter->second).size(); ++i){
res.push_back(strs[(iter->second)[i]]);
}
}
}
return res;
}
};

v2 68ms

class Solution {
public:
//将单词按字母顺序排序后作为hash的key，value等于排序后单词第一次出现大的下标，当第二次出现时value置为-1防止重复再结果中插入第一出现的单词
vector<string> anagrams(vector<string>& strs) {
vector<string> res;
map<string, int> hashTable;
for (int i = 0; i < strs.size(); ++i){
string s = strs[i];
sort(s.begin(), s.end());
map<string,int>::iterator iter = hashTable.find(s);
if (iter != hashTable.end()){
if (iter->second != -1){
res.push_back(strs[iter->second]);
iter->second = -1;
}
res.push_back(strs[i]);
}
else{
hashTable[s] = i;
}
}
return res;
}
};

v3 40ms

class Solution {
public:
const int alphabetList[26] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101 };
vector<string> anagrams(vector<string>& strs) {
vector<string> res;
map<long long, int> hashTable;
for (int i = 0; i < strs.size(); ++i){
long long multi = 1;
for (int j = 0; j < strs[i].size(); ++j){
multi *= alphabetList[strs[i][j]-'a'];
}
map<long long, int>::iterator iter = hashTable.find(multi);
if (iter != hashTable.end()){
if (iter->second != -1){
res.push_back(strs[iter->second]);
iter->second = -1;
}
res.push_back(strs[i]);
}
else{
hashTable[multi] = i;
}
}
return res;
}
};

每一个字母用一个质数代表，用单词中每个字符对应的质数相乘的积作为hash的key（这样查找时不需要比较单词的每一个字符，效率大大提高了），value的意义同上