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Group Anagrams
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Group Anagrams
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/*
LeetCode 49:
Given an array of strings, group anagrams together.
For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],
Return:
[
["ate", "eat","tea"],
["nat","tan"],
["bat"]
]
Note: All inputs will be in lower-case.
Solution:
Language: C++
Tag: hashmap sort
*/
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
if (strs.empty()) {
return vector<vector<string>>();
}
/*
// 1. brute force -- Time Limit Exceed -- O(size) space + O(size^2 len) time
vector<vector<string>> res;
int size = strs.size();
bool *visited = new bool[size];
for (int i = 0; i < size; ++i) {
visited[i] = false;
}
for (int i = 0; i < size; ++i) {
if (visited[i] == true) {
continue;
}
visited[i] = true;
vector<string> group({strs[i]});
for (int j = i + 1; j < size; ++j) {
if (isAnagram(strs[i], strs[j]) == true) {
group.push_back(strs[j]);
visited[j] = true;
}
}
res.push_back(group);
}
return res;
*/
/*
// 2. sort -- O(size * len) space + O(lenloglen * size) time
unordered_map<string, vector<string>> hash;
for (int i = 0; i < strs.size(); ++i) {
string tmp = strs[i];
quicksort(tmp, 0, tmp.size() - 1);
if (hash.find(tmp) == hash.end()) {
vector<string> group({strs[i]});
hash[tmp] = group;
} else {
hash[tmp].push_back(strs[i]);
}
}
vector<vector<string>> res;
for (pair<string, vector<string>> item : hash) {
res.push_back(item.second);
}
return res;
*/
// 3. O(size * len) space + O(len * size) time
unordered_map<string, vector<string>> hashmap;
int *hash = new int[26];
for (int i = 0; i < strs.size(); ++i) {
for (int j = 0; j < 26; ++j) {
hash[j] = 0;
}
for (int j = 0; j < strs[i].length(); ++j) {
++hash[strs[i][j] - 'a'];
}
string tmp = "";
for (int j = 0; j < 26; ++j) {
tmp += ('#' + to_string(hash[j]));
}
if (hashmap.find(tmp) == hashmap.end()) {
vector<string> group({strs[i]});
hashmap[tmp] = group;
} else {
hashmap[tmp].push_back(strs[i]);
}
}
vector<vector<string>> res;
for (pair<string, vector<string>> item : hashmap) {
res.push_back(item.second);
}
return res;
}
bool isAnagram(string &s, string &t) {
if (s.length() != t.length()) {
return false;
}
int *hash = new int[256];
int len = s.length();
for (int i = 0; i < 256; ++i) {
hash[i] = 0;
}
for (int i = 0; i < len; ++i) {
++hash[s[i]];
--hash[t[i]];
}
for (int i = 0; i < 256; ++i) {
if (hash[i] != 0) {
free(hash);
return false;
}
}
free(hash);
return true;
}
void quicksort(string &str, int start, int end) {
if (start >= end) {
return;
}
int left = start, right = end, pivot = str[(start + end) / 2];
while (left <= right) {
while (left <= right && str[left] < pivot) {
++left;
}
while (left <= right && str[right] > pivot) {
--right;
}
if (left <= right) {
swap(str[left++], str[right--]);
}
}
quicksort(str, start, right);
quicksort(str, left, end);
}
};