【並查集】B008_LC_等式方程的可滿足性（互斥）

一、Problem

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.

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.

二、Solution

方法一：UF

class Solution {
	int n, a[];

	void union(int p, int q) {
		int pp = find(p), qq = find(q);
		if (pp == qq) 
			return;
		a[pp] = qq;
	}
	int find(int p) {
		while (a[p] != a[a[p]]) {
			a[p] = a[a[p]];
		}
		return a[p];
	}
    public boolean equationsPossible(String[] es) {
    	n = es.length;
    	a = new int[26];
    	for (int i = 0; i < a.length; i++) a[i] = i;

    	for (String e : es) {
    		for (int i = 0; i < e.length(); i++) {
    			int a = e.charAt(0)-'a', b = e.charAt(3)-'a';
    			if (e.charAt(1) == e.charAt(2))
    				union(a, b);
    		}
    	}
    	for (String e : es) {
    		for (int i = 0; i < e.length(); i++) {
    			int a = e.charAt(0)-'a', b = e.charAt(3)-'a';
    			if (e.charAt(1) != e.charAt(2) && find(a) == find(b))
    				return false;
    		}
    	}
    	return true;
    }
}

小優化：如果有 == 纔去 union，如果有 != 時才比較兩個變量的父親是否相同。

class Solution {
	int n, a[], sz[];

	void union(int p, int q) {
		int pp = find(p), qq = find(q);
		if (pp == qq) 
			return;
        if (sz[pp] > sz[qq]) {
            a[qq] = pp;
            sz[pp] += sz[qq];
        } else {
            a[pp] = qq;
            sz[qq] += sz[pp];
        }
	}
	int find(int p) {
		while (p != a[p]) {
			a[p] = a[a[p]];
            p = a[p];
		}
		return p;
	}
    public boolean equationsPossible(String[] es) {
    	n = es.length;
    	a = new int[26];
        sz = new int[26];
    	for (int i = 0; i < a.length; i++) a[i] = i;

    	for (String e : es)
        for (int i = 0; i < e.length(); i++) if (e.charAt(1) == '=') {
            int a = e.charAt(0)-'a', b = e.charAt(3)-'a';
            if (e.charAt(1) == e.charAt(2))
                union(a, b);
        }
    	for (String e : es)
        for (int i = 0; i < e.length(); i++) if (e.charAt(1) == '!') {
            int a = e.charAt(0)-'a', b = e.charAt(3)-'a';
            if (e.charAt(1) != e.charAt(2) && find(a) == find(b))
                return false;
        }
    	return true;
    }
}

複雜度分析

• 時間複雜度：$O(n × logVarCount)$，
• 空間複雜度：$O(1)$，