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 #

给定一个由表示变量之间关系的字符串方程组成的数组,每个字符串方程 equations[i] 的长度为 4,并采用两种不同的形式之一:“a==b” 或 “a!=b”。在这里,a 和 b 是小写字母(不一定不同),表示单字母变量名。只有当可以将整数分配给变量名,以便满足所有给定的方程时才返回 true,否则返回 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
}


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