LeetCode 990. Satisfiability of Equality Equations

原题链接在这里：https://leetcode.com/problems/satisfiability-of-equality-equations/

题目：

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 '='

题解：

First, go through equations, union all with "==".

Then, go through equations again, if a!=b, but a and b are in the same union, then return false.

Time Complexity: O(nlogk). n = equations.length. k is number distinct character. Since k is single lower case, thus k <= 26.

Space: O(k).

AC Java:

 class Solution {
     HashMap<Character, Character> parent;

     public boolean equationsPossible(String[] equations) {
         if(equations == null || equations.length == 0){
             return true;
         }

         parent = new HashMap<>();
         for(String s : equations){
             if(s.substring(1,3).equals("==")){
                 union(s.charAt(0), s.charAt(3));
             }
         }

         for(String s : equations){
             if(s.substring(1,3).equals("!=") && find(s.charAt(0))==find(s.charAt(3))){
                 return false;
             }
         }

         return true;
     }

     private void union(char i, char j){
         char p = find(i);
         char q = find(j);
         if(p != q){
             parent.put(p, q);
         }
     }

     private char find(char c){
         parent.putIfAbsent(c, c);
         if(parent.get(c) != c){
             char ancestor = find(parent.get(c));
             parent.put(c, ancestor);
         }

         return parent.get(c);
     }
 }