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：

并查集的unoin， find，先合并所有==，之后查找所有!=。

go：

func union(x, y int, f []int) {
    fx := find(x, f)
    fy := find(y, f)
    if fx != fy {
        f[fy] = fx
    }
}

func find(x int, f []int) int{
    if f[x] != x {
        return find(f[x], f)
    } else {
        return x
    }
}

func equationsPossible(equations []string) bool {
    f := make([]int, 26)
    f[0] = 0
    for i := 1; i < 26; i++ {
        f[i] = f[i - 1] + 1
    }
    for _, e := range equations {
        if e[1] == '=' {
            union(int(e[0] - 'a'), int(e[3] - 'a'), f)
        }
    }
    for _, e := range equations {
        if e[1] == '!' {
            if find(int(e[0] - 'a'),f) == find (int(e[3] - 'a'), f) {
                return false
            }
        }
    }
    return true
}