[LeetCode] 990. Satisfiability of Equality Equations

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 '='

等式方程的可满足性。

给定一个由表示变量之间关系的字符串方程组成的数组，每个字符串方程 equations[i] 的长度为 4，并采用两种不同的形式之一："a==b" 或 "a!=b"。在这里，a 和 b 是小写字母（不一定不同），表示单字母变量名。

只有当可以将整数分配给变量名，以便满足所有给定的方程时才返回 true，否则返回 false。 

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/satisfiability-of-equality-equations
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思路是并查集union find。如果中间是等号，说明等号两边的内容应该是同组的，反之则是不同组的。先扫描是等号的，将同组的东西都组合起来；然后扫描不等号的两边，看看是否有跟已经union好的内容有矛盾的地方。

时间O(nlogn)

空间O(n)

Java实现

class Solution {
    public boolean equationsPossible(String[] equations) {
        UF uf = new UF(256);
        for (String e : equations) {
            int letter1 = e.charAt(0) - 'a';
            int letter2 = e.charAt(3) - 'a';
            if (e.charAt(1) == '=') {
                uf.union(letter1, letter2);
            }
        }

        for (String e : equations) {
            if (e.charAt(1) == '!') {
                int letter1 = e.charAt(0) - 'a';
                int letter2 = e.charAt(3) - 'a';
                if (uf.find(letter1) == uf.find(letter2)) {
                    return false;
                }
            }
        }
        return true;
    }
}

class UF {
    int[] parent;
    int count;

    public UF(int n) {
        parent = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
        count = n;
    }

    public int find(int i) {
        if (i == parent[i]) {
            return i;
        }
        return parent[i] = find(parent[i]);
    }

    public void union(int a, int b) {
        int aRoot = find(a);
        int bRoot = find(b);
        if (aRoot != bRoot) {
            parent[aRoot] = bRoot;
            count--;
        }
    }
}

 

LeetCode 题目总结