Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

题意

把由相同字母出现相同次数组成的字符串归到一起(数据可能重复)

题解

一开始我当成了任何一种字母只会出现一次做了,用的是long型作二进制用,自然WA了,其实后来可以用字符串做,当时没想到,看了题解有人将字符串排序作为key,觉得挺有道理的

class Solution {
public:
    struct node {
        string str, after;
        int idx;
        node(string a,string b,int id):str(a),after(b),idx(id){}
        bool operator <(const node&b)const {
            if (after == b.after)
                return idx < b.idx;
            return after < b.after;
        }
    };
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        set<node>all;
        for (int i = 0; i < strs.size(); i++) {
            string tmp = strs[i];
            sort(strs[i].begin(), strs[i].end());
            all.insert(node(tmp, strs[i], i));
            strs[i] = tmp;
        }
        auto pre = all.begin();
        vector<vector<string>>ans;
        vector<string>tmp;
        for (auto p = all.begin(); p != all.end(); p++) {
            if (p != all.begin() && (*p).after != (*pre).after) {
                ans.push_back(tmp);
                tmp.clear();
            }
            tmp.push_back((*p).str);
            pre = p;
        }
        ans.push_back(tmp);
        return ans;
    }
};

View Code

顺便放一下用字符串记录字母出现次数做的,比上面那种慢,因为sort还是挺快的

 1 class Solution {
 2 public:
 3     struct node {
 4         string str, after;
 5         int idx;
 6         node(string a,string b,int id):str(a),after(b),idx(id){}
 7         bool operator <(const node&b)const {
 8             if (after == b.after)
 9                 return idx < b.idx;
10             return after < b.after;
11         }
12     };
13     vector<vector<string>> groupAnagrams(vector<string>& strs) {
14         set<node>all;
15         for (int i = 0; i < strs.size(); i++) {
16             string tmp = "00000000000000000000000000";
17             int l = strs[i].length();
18             for (int j = 0; j < l; j++)
19                 tmp[strs[i][j] - 'a']++;
20             all.insert(node(strs[i], tmp, i));
21         }
22         auto pre = all.begin();
23         vector<vector<string>>ans;
24         vector<string>tmp;
25         for (auto p = all.begin(); p != all.end(); p++) {
26             if (p != all.begin() && (*p).after != (*pre).after) {
27                 ans.push_back(tmp);
28                 tmp.clear();
29             }
30             tmp.push_back((*p).str);
31             pre = p;
32         }
33         ans.push_back(tmp);
34         return ans;
35     }
36 };

View Code