【medium】990. Satisfiability of Equality Equations 并查集

Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b".  Here, a and b are lowercase letters (not necessarily different) that represent one-letter variable names.

Return true if and only if it is possible to assign integers to variable names so as to satisfy all the given equations.

Example 1:

Input: ["a==b","b!=a"]
Output: false
Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the second.  There is no way to assign the variables to satisfy both equations.

Example 2:

Input: ["b==a","a==b"]
Output: true
Explanation: We could assign a = 1 and b = 1 to satisfy both equations.

Example 3:

Input: ["a==b","b==c","a==c"]
Output: true

Example 4:

Input: ["a==b","b!=c","c==a"]
Output: false

Example 5:

Input: ["c==c","b==d","x!=z"]
Output: true

Note:

1. 1 <= equations.length <= 500
2. equations[i].length == 4
3. equations[i][0] and equations[i][3] are lowercase letters
4. equations[i][1] is either '=' or '!'
5. equations[i][2] is '='

 

 

 

class Solution {
private:
    int f[26];
    
    void init(){
        for (int i=0;i<26;i++)
            f[i] = i;
    }
 
    int get_f(int x) {
        if (int(f[x]) == int(x)) {
            return x;
        }
        return f[x] = get_f(f[x]);  // wrong
    }
    
    void merge(int a, int b){
        a = get_f(a);  // wrong
        b = get_f(b);
        f[a] = b;
    }
    
    
    bool is_same_set(int a, int b){
        if (get_f(a) == get_f(b))
            return true;
        return false;
    }
    
public:
    bool equationsPossible(vector<string>& equations) {
        init();
        for (string s : equations){ // first: ==
            cout<<s<<endl;
            int a = s[0] - 'a';  // wrong
            int b = s[3] - 'a';
            if (s[1] == '=')
                merge(a, b);  
        }
         
        for (string s : equations){ // second: !=
            cout<<s<<endl;
            int a = s[0] - 'a';  // wrong
            int b = s[3] - 'a';
            if (s[1] == '!')
                if (is_same_set(a, b))
                    return false;   
        } 
        return true;
    }
};