LeetCode-990. Satisfiability of Equality Equations

LeetCode-990. Satisfiability of Equality Equations

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: falseExplanation: 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: trueExplanation: 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 <= equations.length <= 500​​​​equations[i].length == 4​​​​equations[i][0]​​​ and​​equations[i][3]​​ are lowercase letters​​equations[i][1]​​​ is either​​'='​​​ or​​'!'​​​​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}

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。