#990 Satisfiability of Equality Equations

Description

You are given an array of strings equations that represent relationships between variables where each string equations[i] is of length 4 and takes one of two different forms: “xi==yi” or “xi!=yi”.Here, xi and yi are lowercase letters (not necessarily different) that represent one-letter variable names.

Return true if it is possible to assign integers to variable names so as to satisfy all the given equations, or false otherwise.

Examples

Example 1:

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

Constraints:

1 <= equations.length <= 500
equations[i].length == 4
equations[i][0] is a lowercase letter.
equations[i][1] is either ‘=’ or ‘!’.
equations[i][2] is ‘=’.
equations[i][3] is a lowercase letter.

思路

感觉就是用并查集，判断两个单词是否在同一个簇中，要注意的是，需要先进行 “==” 的union，再进行 “!=” 的判断，因为没有操作来存储 “!=” 的约束

代码

class Solution {public void union(int[] parents, int x, int y) {int px = findParent(parents, x);int py = findParent(parents, y);if (px != py)parents[px] = py;}public int findParent(int[] parents, int x) {if(parents[x] != x)parents[x] = findParent(parents, parents[x]);return parents[x];}public boolean equationsPossible(String[] equations) {int[] parents = new int[26];for (int i = 0; i < parents.length; i++)parents[i] = i;for (String eq: equations) {if (eq.charAt(1) == '=')union(parents, eq.charAt(0) - 'a', eq.charAt(3) - 'a');}for (String eq: equations) {if (eq.charAt(1) == '!') {int x = findParent(parents, eq.charAt(0) - 'a');int y = findParent(parents, eq.charAt(3) - 'a');if (x == y)return false;}}return true;}
}