[Swift]LeetCode886. 可能的二分法 | Possible Bipartition

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➤微信公衆號：山青詠芝（shanqingyongzhi）
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Given a set of N people (numbered 1, 2, ..., N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group. 

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only if it is possible to split everyone into two groups in this way.

Example 1:

Input: N = 4, dislikes = [[1,2],[1,3],[2,4]] Output: true Explanation: group1 [1,4], group2 [2,3] 

Example 2:

Input: N = 3, dislikes = [[1,2],[1,3],[2,3]] Output: false 

Example 3:

Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]] Output: false 

Note:

1. 1 <= N <= 2000
2. 0 <= dislikes.length <= 10000
3. 1 <= dislikes[i][j] <= N
4. dislikes[i][0] < dislikes[i][1]
5. There does not exist i != j for which dislikes[i] == dislikes[j].

---

給定一組 N 人（編號爲 1, 2, ..., N）， 咱們想把每一個人分進任意大小的兩組。

每一個人均可能不喜歡其餘人，那麼他們不該該屬於同一組。

形式上，若是 dislikes[i] = [a, b]，表示不容許將編號爲 a 和 b 的人納入同一組。

當能夠用這種方法將每一個人分進兩組時，返回 true；不然返回 false。 

示例 1：

輸入：N = 4, dislikes = [[1,2],[1,3],[2,4]]
輸出：true
解釋：group1 [1,4], group2 [2,3]

示例 2：

輸入：N = 3, dislikes = [[1,2],[1,3],[2,3]]
輸出：false

示例 3：

輸入：N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
輸出：false 

提示：

1. 1 <= N <= 2000
2. 0 <= dislikes.length <= 10000
3. 1 <= dislikes[i][j] <= N
4. dislikes[i][0] < dislikes[i][1]
5. 對於 dislikes[i] == dislikes[j] 不存在 i != j 

---

【BFS】

Runtime: 836 ms

Memory Usage: 19.8 MB

 1 class Solution {
 2     func possibleBipartition(_ N: Int, _ dislikes: [[Int]]) -> Bool {
 3         var len:Int = dislikes.count
 4         if len < 2 {return true}
 5         var color:[Int] = [Int](repeating:0,count:N + 1)
 6         var graph:[[Int]] = [[Int]](repeating:[Int](),count:N + 1)
 7         for i in 0..<len
 8         {
 9             var m:Int = dislikes[i][0]
10             var n:Int = dislikes[i][1]
11             graph[m].append(n)
12             graph[n].append(m)
13         }
14         for i in 1...N
15         {
16             if color[i] == 0
17             {
18                 color[i] = 1
19                 var q:[Int] = [Int]()
20                 q.append(i)
21                 while (!q.isEmpty)
22                 {
23                     var cur:Int = q.removeFirst()
24                     for j in graph[cur]
25                     {
26                         if color[j] == 0
27                         {
28                             color[j] = color[cur] == 1 ? 2 : 1
29                             q.append(j)
30                         }
31                         else
32                         {
33                             if color[j] == color[cur] {return false}
34                         }
35                     }
36                 }
37             }
38         }
39         return true
40     }
41 }

---

【DFS】

Runtime: 1560 ms

Memory Usage: 50.2 MB

 1 class Solution {
 2     func possibleBipartition(_ N: Int, _ dislikes: [[Int]]) -> Bool {
 3         var graph:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:N),count:N)
 4         for d in dislikes
 5         {
 6             graph[d[0] - 1][d[1] - 1] = 1
 7             graph[d[1] - 1][d[0] - 1] = 1
 8         }
 9         var group:[Int] = [Int](repeating:0,count:N)
10         for i in 0..<N
11         {
12             if group[i] == 0 && !dfs(graph, &group, i, 1)
13             {
14                 return false
15             }
16         }
17         return true
18     }
19 
20     func dfs(_ graph:[[Int]],_ group:inout [Int],_ index:Int,_ g:Int) -> Bool
21     {
22         group[index] = g
23         for i in 0..<graph.count
24         {
25             if graph[index][i] == 1
26             {
27                 if group[i] == g
28                 {
29                     return false
30                 }
31                 if group[i] == 0 && !dfs(graph, &group, i, -g)
32                 {
33                     return false
34                 }
35             }
36         }
37         return true
38     }
39 }