[leetcode]839. Similar String Groups

链接：https://leetcode.com/problems/similar-string-groups/description/

Two strings X and Y are similar if we can swap two letters (in different positions) of X, so that it equals Y.

For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars", "rats", or "arts".

Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}.  Notice that "tars" and "arts"are in the same group even though they are not similar.  Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.

We are given a list A of unique strings.  Every string in A is an anagram of every other string in A.  How many groups are there?

Example 1:

Input: ["tars","rats","arts","star"]
Output: 2

Note:

1. A.length <= 2000
2. A[i].length <= 1000
3. A.length * A[i].length <= 20000
4. All words in A consist of lowercase letters only.
5. All words in A have the same length and are anagrams of each other.
6. The judging time limit has been increased for this question.

思路：

把所以单词连成路径，然后找到不同路径下有多少个终点。

class Solution {
public:
    int numSimilarGroups(vector<string>& A) {
        unordered_map<string,string> mapping;
        int n = A.size();
        for (int i = 0; i < n; i++) {
            /*
            Updated on Jun 6th, there is an invalid test case, in the problem description, it says "a list of unique strings", but there are test cases with duplicate strings.
            In this case, the following line is added.
            */
            if (mapping.find(A[i]) != mapping.end()) continue;
            mapping[A[i]] = A[i];
            for (int j = 0; j < i; j++) {
                if (isSimilar(A[i], A[j]) && mapping[A[j]] != A[i]) {
                    string x = getRoot(A[j], mapping);
                    mapping[x] = A[i];
                }
            }
        }
        
        int result = 0;
        for (auto it = mapping.begin(); it != mapping.end(); it++) {
            if (it->first == it->second) result++;
        }
        return result;
    }
    
private:
    bool isSimilar(string &a, string &b) {
        int n = a.length(), counter = 0;
        for (int i = 0; i < n; i++) {
            if (a[i] != b[i]) counter++;
        }
        return counter == 2;
    }
    
    string getRoot(string s, unordered_map<string,string> &mapping) {
        string temp = mapping[s];
        if (temp != s) return getRoot(temp, mapping);
        else return temp;
    }
};