990. Satisfiability of Equality Equations

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Created At : 2023-10-28 15:14

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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 <= 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 '='

这道题给了一系列简单的公式，某两个字母相等或者不等，然后问给的这些公式会不会产生矛盾，就比如说一个是 a==b，另一个是 a!=b，这就是矛盾了。或者有些更复杂的，相等是可以传递的，比如例子4中，就是一种比较复杂的情况。这里比较直接的一种解法就是建立无向图来做，每个字母都可以当作一个结点，然后等号就表示相连的结点。开始时先跳过所有的不等式，通过所有的等式将这个图建立起来。然后再遍历所有的不等式，看这两个结点在图中是否相连，这里通过递归来检查两个结点是否相连，常规写法，注意要使用一个 HashSet 来标记已经访问过的结点，以免陷入死循环，参见代码如下：

解法一：

class Solution {
public:
    bool equationsPossible(vector<string>& equations) {
        unordered_map<char, unordered_set<char>> g;
        for (auto eq : equations) {
            if (eq[1] == '!') continue;
            g[eq[0]].insert(eq[3]);
            g[eq[3]].insert(eq[0]);
        }
        for (auto eq : equations) {
            if (eq[1] == '=') continue;
            unordered_set<char> visited;
            if (!check(g, eq[0], eq[3], visited)) return false;
        }
        return true;
    }
    bool check(unordered_map<char, unordered_set<char>>& g, char cur, char target, unordered_set<char>& visited) {
        if (cur == target || g[cur].count(target)) return false;
        for (char c : g[cur]) {
            if (visited.count(c)) continue;
            visited.insert(c);
            if (!check(g, c, target, visited)) return false;
        }
        return true;
    }
};

这道题给的提示使用联合查找/并查集 Union Find，论坛上的高分解法也是清一色的 UF 做法。核心思想是初始时给每一个对象都赋上不同的标签，然后对于所有的等式，可以看作是属于同一个群组的对象，需要在 root 数组中标记一下，标记方法是两个对象分别调用 find 函数来找出祖先标签值，然后在 root 数组中再将这两个值连接起来。接下来遍历所有不等式，对不等号两端的对象分别调用 find 函数来找出祖先标签值，若相等了，则产生了矛盾了，因为这两个对象 suppose 不能属于同一个群组的，直接返回 false，参见代码如下：

解法二：

class Solution {
public:
    bool equationsPossible(vector<string>& equations) {
        vector<int> root(26);
        for (int i = 0; i < 26; ++i) root[i] = i;
        for (string eq : equations) {
            if (eq[1] == '!') continue;
            root[find(root, eq[0] - 'a')] = find(root, eq[3] - 'a');
        }
        for (string eq : equations) {
            if (eq[1] == '=') continue;
            if (find(root, eq[0] - 'a') == find(root, eq[3] - 'a')) return false;
        }
        return true;
    }
    int find(vector<int>& root, int x) {
        return root[x] == x ? x : find(root, root[x]);
    }
};

讨论：find 函数的写法其实有很多种，有递归形式，也有迭代形式的，可以参见博主之前的博客 Redundant Connection II。

Github 同步地址:

https://github.com/grandyang/leetcode/issues/990

类似题目：

Redundant Connection II

Friend Circles

Graph Valid Tree

参考资料：

https://leetcode.com/problems/satisfiability-of-equality-equations/

https://leetcode.com/problems/satisfiability-of-equality-equations/discuss/234474/C%2B%2B-7-lines-with-picture-union-find

https://leetcode.com/problems/satisfiability-of-equality-equations/discuss/234486/JavaC%2B%2BPython-Easy-Union-Find

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