[Swift Weekly Contest 123]LeetCode990. 等式方程的可满足性 | Satisfiability of Equality Equations

Given an array equations of strings that represent relationships between variables, each string equations[i] has length 4and 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 '='

---

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。 

示例 1：

输入：["a==b","b!=a"]
输出：false
解释：如果我们指定，a = 1 且 b = 1，那么可以满足第一个方程，但无法满足第二个方程。没有办法分配变量同时满足这两个方程。

示例 2：

输出：["b==a","a==b"]
输入：true
解释：我们可以指定 a = 1 且 b = 1 以满足满足这两个方程。

示例 3：

输入：["a==b","b==c","a==c"]
输出：true

示例 4：

输入：["a==b","b!=c","c==a"]
输出：false

示例 5：

输入：["c==c","b==d","x!=z"]
输出：true

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 '='

---

Runtime: 48 ms

Memory Usage: 4 MB

  1 class Solution {
  2     func equationsPossible(_ equations: [String]) -> Bool {
  3         var ds:DJSet = DJSet(26)
  4         for e in equations
  5         {
  6             if e[1] == "="
  7             {
  8                 var l:Int = e[0].ascii - 97
  9                 var r:Int = e[3].ascii - 97
 10                 ds.union(l,r)
 11             }
 12         }
 13         for e in equations
 14         {
 15             if e[1] == "!"
 16             {
 17                 var l:Int = e[0].ascii - 97
 18                 var r:Int = e[3].ascii - 97
 19                 if ds.equiv(l,r)
 20                 {
 21                     return false
 22                 }
 23             } 
 24         }
 25         return true
 26     }
 27 }
 28 
 29 extension String {        
 30     //subscript函数可以检索数组中的值
 31     //直接按照索引方式截取指定索引的字符
 32     subscript (_ i: Int) -> Character {
 33         //读取字符
 34         get {return self[index(startIndex, offsetBy: i)]}
 35     }
 36 }
 37 
 38 extension Character  
 39 {  
 40   //属性：ASCII整数值(定义小写为整数值)
 41    var ascii: Int {
 42         get {
 43             let s = String(self).unicodeScalars
 44             return Int(s[s.startIndex].value)
 45         }
 46     }
 47 }
 48 
 49 public class DJSet
 50 {
 51     var upper:[Int]
 52     
 53     init(_ n:Int)
 54     {
 55         upper = [Int](repeating:-1,count:n)
 56     }
 57     
 58     func root(_ x:Int) -> Int
 59     {
 60         if(upper[x] < 0)
 61         {
 62             return x
 63         }
 64         else
 65         {
 66             upper[x] = root(upper[x])
 67             return upper[x]
 68         }
 69     }
 70     
 71     func equiv(_ x:Int,_ y:Int) -> Bool
 72     {
 73         return root(x) == root(y)
 74     }
 75     
 76     func union(_ x:Int,_ y:Int) -> Bool
 77     {
 78         var x:Int = root(x)
 79         var y:Int = root(y)
 80         if x != y
 81         {
 82             if upper[y] < upper[x]
 83             {
 84                 var d:Int = x
 85                 x = y
 86                 y = d
 87             }
 88             upper[x]  = upper[y]
 89             upper[y] = x
 90         }
 91         return x == y
 92     }
 93     
 94     func count() -> Int
 95     {
 96         var ct:Int = 0
 97         for u in upper
 98         {
 99             if u < 0
100             {
101                 ct  = 1
102             }
103         }
104         return ct
105     }
106 }

 

来源：http://www./content-4-112251.html