LeetCode——Group Anagrams
Question
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.
解题思路
首先,想到的就是两个for循环一次比较,判断的时候,可以将两个字符串按照字母序排序,来判断两个是否相当,两个for循环,数量多的时候肯定是要超时的。时间复杂度为O(n^2).
所以,这个道题考的是Hash算法,用Map实现,时间复杂度为O(n)。
具体实现
- 一般解法:
#include <iostream>
#include <vector>
#include <map>
using namespace std;
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
if (strs.empty())
return vector<vector<string> >();
int len = strs.size();
vector<vector<string> > ret;
for (int i = 0; i < len; i++)
{
if (strs[i].empty())
continue;
vector<string > sv;
string tmp1 = strs[i];
sv.push_back(tmp1);
sort(tmp1.begin(), tmp1.end());
for (int j = i + 1; j < len; j++)
{
string tmp2 = strs[j];
sort(tmp2.begin(), tmp2.end());
if (tmp1 == tmp2)
{
sv.push_back(strs[j]);
//将处理后的元素赋值为空
strs[j].erase(0, std::string::npos);
}
}//for
//按字典排序该序列
sort(sv.begin(), sv.end());
//添加到结果vector
ret.push_back(sv);
}//for
return ret;
}
};
int main() {
vector<string> vec;
vec.push_back("");
vec.push_back("tea");
vec.push_back("tan");
vec.push_back("ate");
vec.push_back("nat");
vec.push_back("bat");
Solution* solution = new Solution();
vector<vector<string>> result = solution->groupAnagrams(vec);
for (int i = 0; i < result.size(); i++) {
for (int j = 0; j < result[i].size(); j++) {
cout << result[i][j] << " ";
}
cout << endl;
}
return 0;
}
- AC解法:
#include <iostream>
#include <vector>
#include <map>
using namespace std;
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
if (strs.size() == 0)
return vector<vector<string>>();
int size = strs.size();
vector<vector<string>> res;
map<string, vector<string>> str_vec;
for (int i = 0; i < size; i++) {
string str = strs[i];
sort(str.begin(), str.end());
str_vec[str].push_back(strs[i]);
}
for (map<string, vector<string>>::iterator iter = str_vec.begin(); iter != str_vec.end(); iter++) {
sort(iter->second.begin(), iter->second.end());
res.push_back(iter->second);
}
return res;
}
};