【LeetCode & 剑指offer刷题】字符串题14：242. Valid Anagram (变位词系列)
【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
242. Valid Anagram (变位词)
Given two strings s and t , write a function to determine if t is an anagram of s.
Example 1:
Input: s = "anagram", t = "nagaram"
Output: true
Example 2:
Input: s = "rat", t = "car"
Output: false
Note:
You may assume the string contains only lowercase alphabets.
Follow up:
What if the inputs contain unicode characters? How would you adapt your solution to such case?
//问题：字谜/变位词（判断某个字符串是否可以组合成某一个单词，假设只有小写）
/*
方法一：排序之后看是否一样
性能：O(nlogn) O(1)
*/
/*using namespace std;
#include <string>
#include <algorithm>
class Solution
{
public:
bool isAnagram(string s, string t)
{
sort(s.begin(), s.end());
sort(t.begin(), t.end());
return (s==t);
}
};*/
// if(s.find(t) != string::npos) return true; //查找是否有某个子串，不行,因为可能顺序不一样
// else return false;
/*
方法二：统计表法（这里可以直接用数组统计）
延伸：如果key值分布较散，比如包含unicode字符，可以考虑用hash表，
性能：O(n) O(1)
*/
using namespace std;
#include <string>
#include <algorithm>
class Solution
{
public:
bool isAnagram(string s, string t)
{
if(s.size() != t.size()) return false;
int counter[26] = {0}; //26个小写英文字母
for(int i = 0; i<s.size(); i++)
{
counter[s[i]-'a']++;//也可用两个统计表分别统计，最后比较两个统计表是否相等
counter[t[i]-'a']--;
}
for(int count:counter)
{
if(count != 0) return false;
}
return true;
}
};
49. Group Anagrams
Given an array of strings, group anagrams together.
Example:
Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
["ate","eat","tea"],
["nat","tan"],
["bat"]
]
Note:
-
All inputs will be in lowercase.
-
The order of your output does not matter.
//输出字谜组（这里字谜表示通过变序可以组成一个单词的字符串）
//相关题目：Valid Anagram
/*
方法一：用hash表map实现，key--->value为string--->vector<string>
如："aer": ["are", "ear", "era"]
O(nklogk),O(nk)，n为strs的长度，K为strs里字符串的最大长度
*/
#include <unordered_map>
#include <string>
class Solution
{
public:
vector<vector<string>> groupAnagrams(vector<string>& strs)
{
unordered_map<string, vector<string>> mp;
for(string s:strs)
{
string t = s;
sort(t.begin(), t.end()); //key值，字谜单词排序后key值相同
mp[t].push_back(s); //当前key值添加value
}
vector<vector<string>> res;
for(auto& m:mp) //将map中value全部push到结果向量中
{
res.push_back(m.second); //用.second访问map中的键值
}
return res;
}
};
/*
方法二：用统计表法实现，用单词的统计表（可以用vector,长度为26）作为key值
如：[2,1,0,0,...,0]: ["are", "ear", "era"]
O(nk),O(nk)，n为strs的长度，K为strs里字符串的最大长度
general sort takes O(nlogn) time. In this problem, since the string only contains lower-case alphabets, we can write a sorting function using counting sort (O(n) time) to speed up the sorting process.
*/
class Solution {
public:
vector<vector<string>> groupAnagrams(vector<string>& strs) {
unordered_map<string, multiset<string>> mp;
for (string s : strs) {
string t = strSort(s);
mp[t].insert(s);
}
vector<vector<string>> anagrams;
for (auto m : mp) {
vector<string> anagram(m.second.begin(), m.second.end());
anagrams.push_back(anagram);
}
return anagrams;
}
private:
string strSort(string& s) {
int count[26] = {0}, n = s.length();
for (int i = 0; i < n; i++)
count[s[i] - 'a']++; //相当于计数排序
int p = 0;
string t(n, 'a');
for (int j = 0; j < 26; j++) //将“直方图”摊开，j代表某个字母，count[j]代表该字母的个数
for (int i = 0; i < count[j]; i++)
t[p++] += j;
return t;
}
};
