标签：Circles count LeetCode547 return index Int 朋友圈 func var
There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a directfriend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are directfriends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
Example 1:
Input: [[1,1,0], [1,1,0], [0,0,1]] Output: 2 Explanation:The 0th and 1st students are direct friends, so they are in a friend circle.
The 2nd student himself is in a friend circle. So return 2.
Example 2:
Input: [[1,1,0], [1,1,1], [0,1,1]] Output: 1 Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends,
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
Note:
- N is in range [1,200].
- M[i][i] = 1 for all students.
- If M[i][j] = 1, then M[j][i] = 1.
班上有 N 名学生。其中有些人是朋友，有些则不是。他们的友谊具有是传递性。如果已知 A 是 B 的朋友，B 是 C 的朋友，那么我们可以认为 A 也是 C 的朋友。所谓的朋友圈，是指所有朋友的集合。
给定一个 N * N 的矩阵 M，表示班级中学生之间的朋友关系。如果M[i][j] = 1，表示已知第 i 个和 j 个学生互为朋友关系，否则为不知道。你必须输出所有学生中的已知的朋友圈总数。
示例 1:
输入: [[1,1,0], [1,1,0], [0,0,1]] 输出: 2 说明：已知学生0和学生1互为朋友，他们在一个朋友圈。 第2个学生自己在一个朋友圈。所以返回2。
示例 2:
输入: [[1,1,0], [1,1,1], [0,1,1]] 输出: 1 说明：已知学生0和学生1互为朋友，学生1和学生2互为朋友，所以学生0和学生2也是朋友，所以他们三个在一个朋友圈，返回1。
注意：
- N 在[1,200]的范围内。
- 对于所有学生，有M[i][i] = 1。
- 如果有M[i][j] = 1，则有M[j][i] = 1。
180ms
1 class Solution {
2 func findCircleNum(_ M: [[Int]]) -> Int {
3 var count = 0
4 var visited: [Int] = [Int](repeating: 0, count: M.count)
5 for i in 0...(M.count - 1) {
6 if visited[i] == 0 {
7 self.dfs(M, &visited, i)
8 count += 1
9 }
10 }
11
12 return count
13 }
14
15 func dfs(_ M : [[Int]], _ visited: inout [Int], _ i: Int){
16
17 for j in 0...(M.count - 1) {
18 if M[i][j] == 1 && visited[j] == 0 {
19 visited[j] = 1
20 dfs(M, &visited, j)
21 }
22 }
23 }
24 }
Runtime: 184 ms Memory Usage: 19 MB
1 class Solution {
2 //经典思路:联合查找Union Find
3 func findCircleNum(_ M: [[Int]]) -> Int {
4 var n:Int = M.count
5 var res:Int = n
6 var root:[Int] = [Int](repeating:0,count:n)
7 for i in 0..<n
8 {
9 root[i] = i
10 }
11 for i in 0..<n
12 {
13 for j in (i + 1)..<n
14 {
15 if M[i][j] == 1
16 {
17 var p1:Int = getRoot(&root, i)
18 var p2:Int = getRoot(&root, j)
19 if p1 != p2
20 {
21 res -= 1
22 root[p2] = p1
23 }
24 }
25
26 }
27 }
28 return res
29 }
30
31 func getRoot(_ root:inout [Int],_ i:Int) -> Int
32 {
33 var i = i
34 while(i != root[i])
35 {
36 root[i] = root[root[i]]
37 i = root[i]
38 }
39 return i
40 }
41 }
184ms
1 class Solution {
2 func findCircleNum(_ M: [[Int]]) -> Int {
3 var ans = 0
4
5 let size = M.count
6 var visited = [Bool].init(repeating: false, count: size)
7 var current = 0
8 func DFS(_ cur: Int) {
9 guard !visited[cur] else {
10 return
11 }
12 visited[cur] = true
13 for i in 0 ..< size {
14 if !visited[i] && M[cur][i] == 1 {
15 DFS(i)
16 }
17 }
18 }
19 for i in 0 ..< size where !visited[i] {
20 ans += 1
21 DFS(i)
22 }
23 return ans
24 }
25 }
188ms
1 class Solution {
2 func findCircleNum(_ matrix: [[Int]]) -> Int {
3 guard !matrix.isEmpty else {
4 return 0
5 }
6
7 let n = matrix.count
8
9 var visitedStudents = [Bool](repeating: false, count: n)
10 var numOfCircles = 0
11
12 for student in 0..<n {
13 if visitedStudents[student] {
14 continue
15 }
16
17 numOfCircles += 1
18 visitFriends(student: student, matrix: matrix, visitedStudents: &visitedStudents)
19 }
20
21 return numOfCircles
22 }
23
24 func visitFriends(student: Int, matrix: [[Int]], visitedStudents: inout [Bool]) {
25 visitedStudents[student] = true
26
27 for friend in 0..<matrix.count {
28 guard matrix[student][friend] == 1 else {
29 continue
30 }
31 guard !visitedStudents[friend] else {
32 continue
33 }
34
35 visitFriends(student: friend, matrix: matrix, visitedStudents: &visitedStudents)
36 }
37 }
38 }
192ms
1 class Solution {
2 func findCircleNum(_ M: [[Int]]) -> Int {
3 var visited = Set<Int>()
4 var num = 0
5 for i in 0..<M.count {
6 if !visited.contains(i) {
7 self.recurse(i, M, &visited)
8 num += 1
9 }
10 }
11 return num
12 }
13
14 func recurse(_ node: Int, _ graph: [[Int]], _ visited: inout Set<Int>) {
15 visited.insert(node)
16 for i in 0..<graph.count {
17 if !visited.contains(i) && graph[node][i] > 0 {
18 recurse(i, graph, &visited)
19 }
20 }
21 }
22 }
196ms
1 class Solution {
2 func findCircleNum(_ M: [[Int]]) -> Int {
3 guard M.count > 0 && M[0].count > 0 else {
4 return 0
5 }
6
7 var unionFind = UnionFind<Int>()
8
9 for row in 0..<M.count {
10 for column in 0..<M[row].count {
11 if M[row][column] == 1 {
12 unionFind.addSet(row)
13 unionFind.addSet(column)
14 unionFind.union(row, column)
15 }
16 }
17 }
18
19 return unionFind.numerOfSets
20 }
21 }
22
23
24 public struct UnionFind<T: Hashable> {
25 private var index: [T: Int] = [:]
26 private var parent: [Int] = []
27 private var size: [Int] = []
28 public var numerOfSets: Int = 0
29
30 public mutating func addSet(_ element: T) {
31 if index[element] == nil {
32 index[element] = parent.count
33 parent.append(parent.count)
34 size.append(1)
35 numerOfSets += 1
36 }
37 }
38
39 public mutating func find(_ element: T) ->Int? {
40 guard let index = index[element] else {
41 return nil
42 }
43 return setByIndex(index)
44 }
45
46 private mutating func setByIndex(_ index: Int) -> Int {
47 if parent[index] == index {
48 return index
49 } else {
50 parent[index] = setByIndex(parent[index])
51 return parent[index]
52 }
53 }
54
55 public mutating func union(_ element1: T, _ element2: T) {
56 guard let set1 = find(element1), let set2 = find(element2) else {
57 return
58 }
59
60 if set1 != set2 {
61 numerOfSets -= 1
62 if size[set1] > size[set2] {
63 parent[set2] = set1
64 size[set1] += size[set2]
65 } else {
66 parent[set1] = set2
67 size[set2] += size[set1]
68 }
69 }
70 }
71 }
200ms
1 class Solution {
2 class UnionFind {
3 private var parent: [Int]
4 private (set) var count = 0
5
6 init(size: Int, count: Int) {
7 self.parent = Array(0..<size)
8 self.count = count
9 }
10
11 func union(_ x: Int, _ y: Int) {
12 let px = find(x)
13 let py = find(y)
14 if px != py {
15 parent[px] = py
16 count -= 1
17 }
18 }
19
20 func find(_ x: Int) -> Int {
21 if parent[x] == x {
22 return x
23 }
24 parent[x] = find(parent[x])
25 return parent[x]
26 }
27 }
28
29 func findCircleNum(_ M: [[Int]]) -> Int {
30 if M.isEmpty || M[0].isEmpty {
31 return 0
32 }
33
34 let n = M.count
35 let unionFind = UnionFind(size: n, count: n)
36
37 for y in 0..<n {
38 for x in 0..<n {
39 if x != y, M[y][x] == 1 {
40 unionFind.union(x, y)
41 }
42 }
43 }
44 return unionFind.count
45 }
46 }
212ms
1 class UnionFind {
2 private var connectivities: [Int]
3 private var sizes: [Int]
4 init(size: Int) {
5 self.connectivities = [Int](repeating: 0, count: size)
6 self.sizes = [Int](repeating: 1, count: size)
7 for i in 0..<size {
8 connectivities[i] = i
9 }
10 }
11 func connect(n1: Int, n2: Int) {
12 guard let root1 = root(of: n1),
13 let root2 = root(of: n2),
14 root1 != root2 else { return }
15 let sz1 = sizes[root1]
16 let sz2 = sizes[root2]
17 if sz1 > sz2 {
18 connectivities[root2] = root1
19 sizes[root1] += sizes[root2]
20 } else {
21 connectivities[root1] = root2
22 sizes[root2] += sizes[root1]
23 }
24 }
25 func isConnected(n1: Int, n2: Int) -> Bool {
26 return root(of: n1) == root(of: n2)
27 }
28 func root(of n: Int) -> Int? {
29 guard n < connectivities.count else { return nil }
30 let parent = connectivities[n]
31 if parent == n {
32 return n
33 }
34 guard let rt = root(of: parent) else { return nil }
35 connectivities[n] = rt
36 return rt
37 }
38 func snapshot() {
39 print("c: \(connectivities)")
40 print("s: \(sizes)")
41 }
42 }
43
44 class Solution {
45 func findCircleNum(_ M: [[Int]]) -> Int {
46 let size = M.count
47 let uf = UnionFind(size: size)
48 for (y, row) in M.enumerated() {
49 for (x, _) in row.enumerated() {
50 if x <= y {
51 continue
52 }
53 if M[y][x] == 1 {
54 uf.connect(n1: y, n2: x)
55 }
56 }
57 }
58 var groupSet = Set<Int>()
59 for i in 0..<M.count {
60 guard let root = uf.root(of: i) else { continue }
61 groupSet.insert(root)
62 }
63 return groupSet.count
64 }
65 }
216ms
1 class Solution {
2 func findCircleNum(_ M: [[Int]]) -> Int {
3 var visited: Set<Int> = Set<Int>()
4 var res = 0
5 var q: [Int] = []
6 for i in 0..<M.count {
7 if !visited.contains(i) {
8 res += 1
9 q.append(i)
10 while !q.isEmpty {
11 let friend = q.removeFirst()
12 if !visited.contains(friend) {
13 visited.insert(friend)
14 for j in 0..<M[friend].count {
15 if M[friend][j] == 1 {
16 q.append(j)
17 }
18 }
19 }
20 }
21 }
22 }
23 return res
24 }
25 }
228ms
1 class Solution {
2 func findCircleNum(_ M: [[Int]]) -> Int {
3
4 var sum = 0
5 var circle = M
6 let mCount = M.count
7 for i in 0 ..< mCount {
8 let isCircle = p_circle(&circle, down: i, right: i, s: mCount)
9 if isCircle { sum += 1 }
10 }
11 return sum
12 }
13
14 func p_circle(_ circle: inout [[Int]], down: Int, right: Int, s: Int) -> Bool {
15 guard
16 down < s,
17 right <= down,
18 circle[down][right] == 1
19 else { return false }
20 circle[down][right] = 0
21 circle[right][down] = 0
22
23 var r = 0
24 if down == 0 { r = 1 }
25 while (r < s) {
26 if circle[down][r] == 1 {
27 _ = p_circle(&circle, down: r, right: r, s: s)
28 }
29 r += 1
30 if down == r { r += 1 }
31 }
32 return true
33 }
34 }
328ms
1 class FriendCircle {
2 var friends: Set<Int> = []
3 }
4
5 extension FriendCircle: Hashable {
6 var hashValue: Int {
7 return ObjectIdentifier(self).hashValue
8 }
9
10 static func ==(lhs: FriendCircle, rhs: FriendCircle) -> Bool {
11 return lhs === rhs
12 }
13 }
14
15
16 class Solution {
17 func findCircleNum(_ matrix: [[Int]]) -> Int {
18 guard !matrix.isEmpty else {
19 return 0
20 }
21
22 let n = matrix.count
23 var numOfCircles = 0
24 var personToCircle = [FriendCircle?](repeating: nil, count: n)
25
26 for column in 0..<n {
27
28 var friends: Set<Int> = []
29 var circles: Set<FriendCircle> = []
30
31 for row in 0..<n {
32 if matrix[row][column] == 1 {
33 friends.insert(row)
34
35 if let existingCircle = personToCircle[row] {
36 circles.insert(existingCircle)
37 }
38 }
39 }
40
41 let circle: FriendCircle
42 if circles.count > 1 {
43 for circle in circles {
44 friends.formUnion(circle.friends)
45 }
46 circle = FriendCircle()
47 numOfCircles -= circles.count-1
48 } else if let existingCircle = circles.first {
49 circle = existingCircle
50 } else {
51 circle = FriendCircle()
52 numOfCircles += 1
53 }
54
55
56 for friend in friends {
57 circle.friends.insert(friend)
58 personToCircle[friend] = circle
59 }
60 }
61 return numOfCircles
62 }
63 }
352ms
1 class Solution {
2
3 private var circles: [Int] = []
4 private var count = 0
5
6 func connect(_ index: Int, _ index2: Int) {
7 let af = find(index)
8 let bf = find(index2)
9 guard af != bf else {
10 return
11 }
12
13 for index in 0..<circles.count {
14 if circles[index] == af {
15 circles[index] = bf
16 }
17 }
18
19 count -= 1
20 }
21
22 func find(_ index: Int) -> Int {
23 if index >= 0 && index < circles.count {
24 return circles[index]
25 } else {
26 return -1
27 }
28 }
29
30 func findCircleNum(_ M: [[Int]]) -> Int {
31 guard !M.isEmpty && !M[0].isEmpty else {
32 return 0
33 }
34
35 count = M.count
36 for index in 0..<M.count {
37 circles.append(index)
38 }
39
40 for row in 0..<M.count {
41 for column in 0..<M[row].count {
42 if row != column && M[row][column] == 1 {
43 connect(row, column)
44 }
45 }
46 }
47
48 return count
49 }
50 }
384ms
1 class Solution {
2 var pre:[Int]
3 init() {
4 self.pre = [Int]()
5 }
6 func findCircleNum(_ M: [[Int]]) -> Int {
7
8 if (M.count > 0) {
9 self.pre = [Int](repeating: 0, count: M.count)
10 for (i, v) in pre.enumerated() {
11 pre[i] = i
12 }
13 for i in 0...M.count - 1 {
14 if (M[i].count == M.count) {
15 for j in 0...M[i].count - 1 {
16 // print("a loop 1")
17 if (M[i][j] == 1) {
18 join(i, j)
19 }
20 }
21 } else {
22 return 0 //异常数据
23 }
24 }
25 var count = 0
26 for (i, v) in pre.enumerated() {
27 if (i == v) {
28 count+=1
29 }
30 }
31 // print("\(count)")
32 return count
33 }
34 return 0
35 }
36
37 func join( _ i:Int, _ j:Int) {
38 var preI = find(i)
39 var preJ = find(j)
40 if (preI != preJ) {
41 self.pre[preI] = preJ
42 }
43 // print("\(self.pre)")
44 }
45 func find(_ i:Int)->Int {
46 var i = i
47 while (self.pre[i] != i) {
48 i = self.pre[i]
49 }
50 return i
51 }
52 }
400ms
1 class Solution {
2 func findCircleNum(_ M: [[Int]]) -> Int {
3 var ret = 0
4 var visited:Array<Bool> = Array(repeating: false, count: M.count)
5 func getFriend(_ y:Int) {
6 for x in 0..<M.count {
7 if !visited[x] && M[y][x] == 1{
8 visited[y] = true
9 getFriend(x)
10 }
11 }
12 }
13 guard M.count > 0 else {
14 return ret
15 }
16 for y in 0..<M.count {
17 if(!visited[y]) {
18 ret += 1
19 getFriend(y)
20 }
21 }
22 return ret
23 }
24 }
标签：Circles,count,LeetCode547,return,index,Int,朋友圈,func,var 来源： https://www.cnblogs.com/strengthen/p/10414531.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。