題目

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 '='

題目大意

給定一個由表示變量之間關係的字符串方程組成的數組，每個字符串方程 equations[i] 的長度為 4，並採用兩種不同的形式之一："a==b" 或 "a!=b"。在這裏，a 和 b 是小寫字母（不一定不同），表示單字母變量名。只有當可以將整數分配給變量名，以便滿足所有給定的方程時才返回 true，否則返回 false。

提示：

1. 1 <= equations.length <= 500
2. equations[i].length == 4
3. equations[i][0] 和 equations[i][3] 是小寫字母
4. equations[i][1] 要麼是 '='，要麼是 '!'
5. equations[i][2] 是 '='

解題思路

• 給出一個字符串數組，數組裏面給出的是一些字母的關係，只有 '==' 和 '! =' 兩種關係。問給出的這些關係中是否存在悖論？
• 這一題是簡單的並查集的問題。先將所有 '==' 關係的字母 union() 起來，然後再一一查看 '! =' 關係中是否有 '==' 關係的組合，如果有，就返回 false，如果遍歷完都沒有找到，則返回 true。 e`。

參考代碼

package leetcode

import (
	"github.com/halfrost/LeetCode-Go/template"
)

func equationsPossible(equations []string) bool {
	if len(equations) == 0 {
		return false
	}
	uf := template.UnionFind{}
	uf.Init(26)
	for _, equ := range equations {
		if equ[1] == '=' && equ[2] == '=' {
			uf.Union(int(equ[0]-'a'), int(equ[3]-'a'))
		}
	}
	for _, equ := range equations {
		if equ[1] == '!' && equ[2] == '=' {
			if uf.Find(int(equ[0]-'a')) == uf.Find(int(equ[3]-'a')) {
				return false
			}
		}
	}
	return true
}