990. Satisfiability of Equality Equations

问题：

给定多个字符串定义：某个小写字母==（！=）另一个小写字母

求这些定义中，有互相矛盾的情况吗？

有矛盾，返回false，否则返回true。

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

　　

解法：并查集（Disjoint Set）

先遍历一遍所有字符串定义，将==连接的两个小写字母 连接。

再遍历一次所有字符串定义，对于!=两边的两个小写字母，检测是否已经连接，

若已经连接，则说明矛盾，返回false。

全部遍历完毕，都没有返回，则证明全部不矛盾，返回true。

代码参考：

class Solution {
public:
    bool equationsPossible(vector<string>& equations) {
        DisjointSet ds(26);
        for(string& eq:equations) {
            if(eq[1]=='=') {
                ds.merge(eq[0]-'a', eq[3]-'a');
            }
        }
        for(string& eq:equations) {
            if(eq[1]=='!') {
                if(ds.isConnected(eq[0]-'a', eq[3]-'a')) return false;
            }
        }
        return true;
    }
};

并查集：

class DisjointSet {
public:
    DisjointSet(int n):root(n, 0), rank(n, 0) {
        for(int i=0; i<n; i++) root[i]=i;
    }
    int find(int i) {
        if(i!=root[i]) {
            root[i] = find(root[i]);
        }
        return root[i];
    }
    void merge(int x, int y) {
        int x_root = find(x);
        int y_root = find(y);
        if(x_root==y_root) return;
        if(rank[x_root]>rank[y_root]) {
            root[y_root] = x_root;
        } else if(rank[x_root]<rank[y_root]) {
            root[x_root] = y_root;
        } else {
            root[y_root] = x_root;
            rank[x_root] ++;
        }
        return;
    }
    bool isConnected(int x, int y) {
        int x_root = find(x);
        int y_root = find(y);
        return x_root == y_root;
    }
private:
    vector<int> root;
    vector<int> rank;
};