[Google] LeetCode 839 Similar String Groups 并查集

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

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 strs of strings where every string in strs is an anagram of every other string in strs. How many groups are there?

Solution

可以交换两个字符的可以归为一类，所以我们可以维护一个并查集，当然此时我们用 \(unordered\_map\) 来维护父节点. 最后只需要统计多少个

\[fa[x]=x \]

点击查看代码

class Solution {
private:
    bool check(string& a, string& b){
        int cnt=0;
        int l = a.size();
        for(int i=0;i<l;i++){
            cnt += (a[i]!=b[i]);
        }
        return cnt==0 || cnt==2;
    }
    
    unordered_map<string,string> fa;
    unordered_map<string,int> sz;
    
    string find(string s, unordered_map<string,string>& fa){
        if(s==fa[s])return s;
        return fa[s] = find(fa[s], fa);
    }
    
    void join(string u, string v,unordered_map<string,string>&fa, unordered_map<string,int>& sz){
        u = find(u, fa); v = find(v, fa);
        if(sz[u]>sz[v])fa[v]=u;
        else if(sz[v]>sz[u])fa[u]=v;
        else {fa[v]=u;sz[u]++;}
    }
    
    int ans=0;
public:
    int numSimilarGroups(vector<string>& strs) {
        int n = strs.size();
        int l = strs[0].size();
        for(int i=0;i<n;i++){
            fa[strs[i]] = strs[i];
            sz[strs[i]]=0;
        }
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(check(strs[i], strs[j])){
                    join(strs[i], strs[j], fa, sz);
                }
            }
        }
        for(auto ele: fa){
            if(ele.first ==ele.second)ans++;
        }
        return ans;
    }
};