0990. Satisfiability of Equality Equations

# 990. 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 `'='`

## 题目大意 #

1. 1 <= equations.length <= 500
2. equations[i].length == 4
3. equations[i][0] 和 equations[i][3] 是小写字母
4. equations[i][1] 要么是 ‘=’，要么是 ‘!’
5. equations[i][2] 是 ‘=’

## 解题思路 #

• 给出一个字符串数组，数组里面给出的是一些字母的关系，只有 `'=='``'! ='` 两种关系。问给出的这些关系中是否存在悖论？
• 这一题是简单的并查集的问题。先将所有 `'=='` 关系的字母 `union()` 起来，然后再一一查看 `'! ='` 关系中是否有 `'=='` 关系的组合，如果有，就返回 `false`，如果遍历完都没有找到，则返回 `true`

## 代码 #

``````
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
}

``````

Sep 6, 2020