Given an undirected graph, return true if and only if it is bipartite.
Recall that a graph is bipartite if we can split it's set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.
The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn't contain any element twice.
Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation:
The graph looks like this:
0----1
| |
| |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.
Example 2: Input: [[1,2,3], [0,2], [0,1,3], [0,2]] Output: false Explanation: The graph looks like this: 0----1 | \ | | \ | 3----2 We cannot find a way to divide the set of nodes into two independent subsets.
Note:
graphwill have length in range[1, 100].graph[i]will contain integers in range[0, graph.length - 1].graph[i]will not containior duplicate values.- The graph is undirected: if any element
jis ingraph[i], theniwill be ingraph[j].
给定一个无向图graph，当这个图为二分图时返回true。
如果我们能将一个图的节点集合分割成两个独立的子集A和B，并使图中的每一条边的两个节点一个来自A集合，一个来自B集合，我们就将这个图称为二分图。
graph将会以邻接表方式给出，graph[i]表示图中与节点i相连的所有节点。每个节点都是一个在0到graph.length-1之间的整数。这图中没有自环和平行边： graph[i] 中不存在i，并且graph[i]中没有重复的值。
示例 1:
输入: [[1,3], [0,2], [1,3], [0,2]]
输出: true
解释:
无向图如下:
0----1
| |
| |
3----2
我们可以将节点分成两组: {0, 2} 和 {1, 3}。
示例 2:
输入: [[1,2,3], [0,2], [0,1,3], [0,2]]
输出: false
解释:
无向图如下:
0----1
| \ |
| \ |
3----2
我们不能将节点分割成两个独立的子集。
注意:
graph的长度范围为[1, 100]。graph[i]中的元素的范围为[0, graph.length - 1]。graph[i]不会包含i或者有重复的值。- 图是无向的: 如果
j在graph[i]里边, 那么i也会在graph[j]里边。
Runtime: 176 ms
Memory Usage: 19.2 MB
1 class Solution { 2 func isBipartite(_ graph: [[Int]]) -> Bool { 3 var colors:[Int] = [Int](repeating:0,count:graph.count) 4 for i in 0..<graph.count 5 { 6 if colors[i] != 0 {continue} 7 colors[i] = 1 8 var q:[Int] = [i] 9 while(!q.isEmpty) 10 { 11 var t:Int = q.first! 12 q.removeFirst() 13 for a in graph[t] 14 { 15 if colors[a] == colors[t] {return false} 16 if colors[a] == 0 17 { 18 colors[a] = -1 * colors[t] 19 q.append(a) 20 } 21 } 22 } 23 } 24 return true 25 } 26 }
208ms
1 class Solution { 2 func isBipartite(_ graph: [[Int]]) -> Bool { 3 let V = graph.count 4 var colors = Array(repeating: -1, count: V) 5 for i in 0 ..< V { 6 if colors[i] == -1 && !validColor(&colors, 0, graph, i) { 7 return false 8 } 9 } 10 return true 11 } 12 13 func validColor(_ colors: inout [Int], _ color: Int, _ graph: [[Int]], _ vertex: Int) -> Bool { 14 if colors[vertex] != -1 { 15 return colors[vertex] == color 16 } 17 colors[vertex] = color 18 for neighbor in graph[vertex] { 19 if !validColor(&colors, 1 - color, graph, neighbor) { 20 return false 21 } 22 } 23 return true 24 } 25 }
212ms
1 class Solution { 2 func isBipartite(_ graph: [[Int]]) -> Bool { 3 var color = Array<Color>(repeating: .unknown, count: graph.count) 4 5 for i in 0..<graph.count { 6 guard color[i] == .unknown else { continue } 7 if !dfs(i: i, color: &color, graph: graph, paint: .red) { return false } 8 } 9 return true 10 } 11 func dfs(i: Int, color: inout [Color], graph: [[Int]], paint: Color) -> Bool { 12 guard color[i] == .unknown else { 13 return color[i] == paint 14 } 15 color[i] = paint 16 for neighbour in graph[i] { 17 if !dfs(i: neighbour, color: &color, graph: graph, paint: paint.opposite) 18 { 19 return false 20 } 21 } 22 return true 23 } 24 } 25 26 enum Color { 27 case red 28 case blue 29 case unknown 30 31 var opposite: Color { 32 switch self { 33 case .red: 34 return .blue 35 case .blue: 36 return .red 37 case .unknown: 38 return .unknown 39 } 40 } 41 }
220ms
1 class Solution { 2 func isBipartite(_ graph: [[Int]]) -> Bool { 3 var visited: Set<Int> = [] 4 5 for seed in 0..<graph.count { 6 if visited.contains(seed) { continue } 7 8 var oddSet: Set<Int> = [] 9 var evenSet: Set<Int> = [] 10 11 var queue = [seed] 12 var level = 0 13 14 while !queue.isEmpty { 15 var nextLevel = [Int]() 16 17 for vertex in queue { 18 visited.insert(vertex) 19 if level % 2 == 0 { 20 if oddSet.contains(vertex) { 21 return false 22 } 23 24 if !evenSet.contains(vertex) { 25 nextLevel += graph[vertex] 26 evenSet.insert(vertex) 27 } 28 } else { 29 if evenSet.contains(vertex) { 30 return false 31 } 32 33 if !oddSet.contains(vertex) { 34 nextLevel += graph[vertex] 35 oddSet.insert(vertex) 36 } 37 } 38 } 39 queue = nextLevel 40 level += 1 41 } 42 } 43 return true 44 } 45 }
224ms
1 class Solution { 2 enum NodeStatus { 3 case red, black, undefined 4 func negate() -> NodeStatus { 5 switch self { 6 case .red: 7 return .black 8 case .black: 9 return .red 10 case .undefined: 11 return .undefined 12 } 13 } 14 } 15 16 var nodeStatus: [Int: NodeStatus] = [:] 17 var graph: [[Int]] = [] 18 var nodeCount = 0 19 20 func status(_ node: Int) -> NodeStatus { 21 return nodeStatus[node] ?? .undefined 22 } 23 24 func findConflict(byFlooding nodes: [Int]) -> Bool { 25 var nextNodes = nodes 26 while nextNodes.count > 0 { 27 let currentNodes = nextNodes 28 nextNodes = [Int]() 29 for node in currentNodes { 30 for adjacent in graph[node] { 31 let currentStatus = status(node) 32 let statusToColor = currentStatus.negate() 33 if nodeStatus[adjacent] == .undefined { 34 nodeStatus[adjacent] = statusToColor 35 nextNodes.append(adjacent) 36 } else if nodeStatus[adjacent] == currentStatus { 37 return true 38 } 39 } 40 } 41 } 42 return false 43 } 44 45 func undefinedNode() -> Int? { 46 for i in 0..<nodeCount { 47 if status(i) == .undefined { 48 return i 49 } 50 } 51 return nil 52 } 53 54 func setup(_ graph: [[Int]]) { 55 self.graph = graph 56 nodeCount = graph.count 57 for i in 0..<graph.count { 58 nodeStatus[i] = .undefined 59 } 60 } 61 62 func isBipartite(_ graph: [[Int]]) -> Bool { 63 setup(graph) 64 while let node = undefinedNode() { 65 nodeStatus[node] = .red 66 let hasConflict = findConflict(byFlooding: [node]) 67 if hasConflict { 68 return false 69 } 70 } 71 72 return true 73 } 74 }
228ms
1 class Solution { 2 func isBipartite(_ graph: [[Int]]) -> Bool { 3 guard graph.count > 0 else { 4 return false 5 } 6 let black = 0 7 let white = 1 8 var visited = [Int:Int]() 9 10 func bfs(node:Int, pColor:Int) -> Bool { 11 if let color = visited[node] { 12 return color != pColor 13 } 14 15 visited[node] = pColor == white ? black : white 16 17 let neighb = graph[node] 18 for n in neighb { 19 if !bfs(node:n, pColor:visited[node]!) { 20 return false 21 } 22 } 23 return true 24 } 25 26 for i in 0..<graph.count { 27 if visited[i] == nil { 28 if !bfs(node:i, pColor:white) { 29 return false 30 } 31 } 32 } 33 34 return true 35 } 36 }