In this problem, a tree is an undirected graph that is connected and has no cycles.
The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., N), with one additional edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.
The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.
Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.
Example 1:
Input: [[1,2], [1,3], [2,3]] Output: [2,3] Explanation: The given undirected graph will be like this: 1 / 2 - 3
Example 2:
Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
| |
4 - 3
Note:
- The size of the input 2D-array will be between 3 and 1000.
- Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.
Update (2017-09-26):
We have overhauled the problem description + test cases and specified clearly the graph is an undirectedgraph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.
在本问题中, 树指的是一个连通且无环的无向图。
输入一个图，该图由一个有着N个节点 (节点值不重复1, 2, ..., N) 的树及一条附加的边构成。附加的边的两个顶点包含在1到N中间，这条附加的边不属于树中已存在的边。
结果图是一个以边组成的二维数组。每一个边的元素是一对[u, v] ，满足 u < v，表示连接顶点u 和v的无向图的边。
返回一条可以删去的边，使得结果图是一个有着N个节点的树。如果有多个答案，则返回二维数组中最后出现的边。答案边 [u, v] 应满足相同的格式 u < v。
示例 1：
输入: [[1,2], [1,3], [2,3]] 输出: [2,3] 解释: 给定的无向图为: 1 / 2 - 3
示例 2：
输入: [[1,2], [2,3], [3,4], [1,4], [1,5]]
输出: [1,4]
解释: 给定的无向图为:
5 - 1 - 2
| |
4 - 3
注意:
- 输入的二维数组大小在 3 到 1000。
- 二维数组中的整数在1到N之间，其中N是输入数组的大小。
更新(2017-09-26):
我们已经重新检查了问题描述及测试用例，明确图是无向 图。对于有向图详见冗余连接II。对于造成任何不便，我们深感歉意。
Runtime: 28 ms
Memory Usage: 18.9 MB
1 class Solution {
2 func findRedundantConnection(_ edges: [[Int]]) -> [Int] {
3 var root:[Int] = [Int](repeating:-1,count:2001)
4 for edge in edges
5 {
6 var x:Int = find(&root, edge[0])
7 var y:Int = find(&root, edge[1])
8 if x == y {return edge}
9 root[x] = y
10 }
11 return [Int]()
12 }
13
14 func find (_ root:inout [Int],_ i:Int) -> Int
15 {
16 var i = i
17 while (root[i] != -1)
18 {
19 i = root[i]
20 }
21 return i
22 }
23 }
32ms
1 class Solution {
2 func findRedundantConnection(_ edges: [[Int]]) -> [Int] {
3 guard edges.isEmpty == false else {
4 return []
5 }
6 let n = edges.count
7 var parents: [Int] = []
8 for i in 0...n {
9 parents.append(i)
10 }
11 for edge in edges {
12 let first = edge[0]
13 let second = edge[1]
14 let p1 = find(parents, first)
15 let p2 = find(parents, second)
16 if p1 == p2 {
17 return edge
18 }
19 parents[p2] = p1
20 }
21 return []
22 }
23
24 private func find(_ parents: [Int], _ val: Int) -> Int {
25 if parents[val] == val {
26 return val
27 }
28 return find(parents, parents[val])
29 }
30 }
48ms
1 class Solution {
2 // s1: union find
3 func findRedundantConnection(_ edges: [[Int]]) -> [Int] {
4 var uf = UnionFind(n: edges.count+1)
5 for con in edges {
6 let s = con[0]
7 let e = con[1]
8 if uf.union(s, e) == false {
9 return con
10 }
11 }
12 return []
13 }
14 }
15
16 class UnionFind {
17 public var parents = [Int]()
18 private var ranks = [Int]()
19 public var count: Int = 0
20 init(n: Int) {
21 for i in 0..<n {
22 parents.append(i)
23 ranks.append(1)
24 }
25 }
26
27 func find(_ x: Int) -> Int {
28 var x = x
29 if parents[x] != x {
30 parents[x] = find(parents[x])
31 }
32 return parents[x]
33 }
34 /*
35 1 2 3
36 5 6
37 */
38 func union(_ x: Int, _ y: Int) -> Bool {
39 let px = find(x)
40 let py = find(y)
41 if px == py {
42 return false
43 }
44 count -= 1
45 if ranks[x] > ranks[y] {
46 parents[py] = px
47 } else if ranks[x] < ranks[y] {
48 parents[px] = py
49 } else {
50 parents[py] = px
51 ranks[px] += 1
52 }
53 return true
54 }
55 }
52ms
1 class Solution {
2 func findRedundantConnection(_ edges: [[Int]]) -> [Int] {
3 guard edges.count > 0 else { return [0,0] }
4
5 var totalNode = edges.count + 1
6
7 var group: [Int] = []
8 var groupLevel: [Int] = []
9
10 for i in 0..<totalNode {
11 group.append(i)
12 groupLevel.append(0)
13 }
14
15 var extraEdge:[Int] = []
16
17 for edge in edges {
18 var nodeX = edge[0]
19 var nodeY = edge[1]
20
21 var pNodeX = findParent(nodeX, &group)
22 var pNodeY = findParent(nodeY, &group)
23 if pNodeX != pNodeY {
24 if groupLevel[pNodeX] > groupLevel[pNodeY] {
25 group[pNodeY] = pNodeX
26 }else if groupLevel[pNodeX] < groupLevel[pNodeY] {
27 group[pNodeX] = pNodeY
28 }else {
29 group[pNodeY] = pNodeX
30 groupLevel[pNodeX] += 1
31 }
32 }else {
33 extraEdge = edge
34 }
35 }
36 return extraEdge
37 }
38
39
40 func findParent(_ node: Int, _ group: inout [Int]) -> Int {
41 var currentNode = node
42 while currentNode != group[currentNode] {
43 group[currentNode] = group[group[currentNode]]
44 currentNode = group[currentNode]
45 }
46
47 return currentNode
48 }
49 }
64ms
1 class Solution {
2
3 struct Edge {
4 var w: Int
5 var a: Int
6 var b: Int
7 }
8
9 func findRedundantConnection(_ _edges: [[Int]]) -> [Int] {
10 var wEdges = [Int: [(Int, Int)]]()
11 for (i, edge) in _edges.enumerated() {
12 wEdges[edge[0], default: []].append((edge[1], i))
13 wEdges[edge[1], default: []].append((edge[0], i))
14 }
15 var safe: Set<Int> = []
16 var heap = Heap<((Int, Int), Int)>(sort: {
17 $0.1 < $1.1
18 })
19 let source = _edges[0][0]
20 var edges = Set<[Int]>()
21 safe.insert(source)
22 for n in wEdges[source]! {
23 heap.insert( ((source, n.0), n.1) )
24 }
25 while !heap.isEmpty {
26 let ((source, node), _) = heap.remove()!
27 safe.insert(node)
28 edges.insert([source, node])
29 edges.insert([node, source])
30 for n in wEdges[node]! {
31 if edges.contains( [n.0, node] ) {
32
33 } else if safe.contains(n.0) {
34 return [node, n.0].sorted()
35 } else {
36 heap.insert( ((node, n.0), n.1) )
37 }
38 }
39 }
40
41 return _edges.last!
42 }
43 }
44
45
46
47
48
49 public struct Heap<T> {
50
51 /** The array that stores the heap‘s nodes. */
52 var nodes = [T]()
53
54 /**
55 * Determines how to compare two nodes in the heap.
56 * Use ‘>‘ for a max-heap or ‘<‘ for a min-heap,
57 * or provide a comparing method if the heap is made
58 * of custom elements, for example tuples.
59 */
60 private var orderCriteria: (T, T) -> Bool
61
62 /**
63 * Creates an empty heap.
64 * The sort function determines whether this is a min-heap or max-heap.
65 * For comparable data types, > makes a max-heap, < makes a min-heap.
66 */
67 public init(sort: @escaping (T, T) -> Bool) {
68 self.orderCriteria = sort
69 }
70
71 /**
72 * Creates a heap from an array. The order of the array does not matter;
73 * the elements are inserted into the heap in the order determined by the
74 * sort function. For comparable data types, ‘>‘ makes a max-heap,
75 * ‘<‘ makes a min-heap.
76 */
77 public init(array: [T], sort: @escaping (T, T) -> Bool) {
78 self.orderCriteria = sort
79 configureHeap(from: array)
80 }
81
82 /**
83 * Configures the max-heap or min-heap from an array, in a bottom-up manner.
84 * Performance: This runs pretty much in O(n).
85 */
86 private mutating func configureHeap(from array: [T]) {
87 nodes = array
88 for i in stride(from: (nodes.count/2-1), through: 0, by: -1) {
89 shiftDown(i)
90 }
91 }
92
93 public var isEmpty: Bool {
94 return nodes.isEmpty
95 }
96
97 public var count: Int {
98 return nodes.count
99 }
100
101 /**
102 * Returns the index of the parent of the element at index i.
103 * The element at index 0 is the root of the tree and has no parent.
104 */
105 @inline(__always) internal func parentIndex(ofIndex i: Int) -> Int {
106 return (i - 1) / 2
107 }
108
109 /**
110 * Returns the index of the left child of the element at index i.
111 * Note that this index can be greater than the heap size, in which case
112 * there is no left child.
113 */
114 @inline(__always) internal func leftChildIndex(ofIndex i: Int) -> Int {
115 return 2*i + 1
116 }
117
118 /**
119 * Returns the index of the right child of the element at index i.
120 * Note that this index can be greater than the heap size, in which case
121 * there is no right child.
122 */
123 @inline(__always) internal func rightChildIndex(ofIndex i: Int) -> Int {
124 return 2*i + 2
125 }
126
127 /**
128 * Returns the maximum value in the heap (for a max-heap) or the minimum
129 * value (for a min-heap).
130 */
131 public func peek() -> T? {
132 return nodes.first
133 }
134
135 /**
136 * Adds a new value to the heap. This reorders the heap so that the max-heap
137 * or min-heap property still holds. Performance: O(log n).
138 */
139 public mutating func insert(_ value: T) {
140 nodes.append(value)
141 shiftUp(nodes.count - 1)
142 }
143
144 /**
145 * Adds a sequence of values to the heap. This reorders the heap so that
146 * the max-heap or min-heap property still holds. Performance: O(log n).
147 */
148 public mutating func insert<S: Sequence>(_ sequence: S) where S.Iterator.Element == T {
149 for value in sequence {
150 insert(value)
151 }
152 }
153
154 /**
155 * Allows you to change an element. This reorders the heap so that
156 * the max-heap or min-heap property still holds.
157 */
158 public mutating func replace(index i: Int, value: T) {
159 guard i < nodes.count else { return }
160
161 remove(at: i)
162 insert(value)
163 }
164
165 /**
166 * Removes the root node from the heap. For a max-heap, this is the maximum
167 * value; for a min-heap it is the minimum value. Performance: O(log n).
168 */
169 @discardableResult public mutating func remove() -> T? {
170 guard !nodes.isEmpty else { return nil }
171
172 if nodes.count == 1 {
173 return nodes.removeLast()
174 } else {
175 // Use the last node to replace the first one, then fix the heap by
176 // shifting this new first node into its proper position.
177 let value = nodes[0]
178 nodes[0] = nodes.removeLast()
179 shiftDown(0)
180 return value
181 }
182 }
183
184 /**
185 * Removes an arbitrary node from the heap. Performance: O(log n).
186 * Note that you need to know the node‘s index.
187 */
188 @discardableResult public mutating func remove(at index: Int) -> T? {
189 guard index < nodes.count else { return nil }
190
191 let size = nodes.count - 1
192 if index != size {
193 nodes.swapAt(index, size)
194 shiftDown(from: index, until: size)
195 shiftUp(index)
196 }
197 return nodes.removeLast()
198 }
199
200 /**
201 * Takes a child node and looks at its parents; if a parent is not larger
202 * (max-heap) or not smaller (min-heap) than the child, we exchange them.
203 */
204 internal mutating func shiftUp(_ index: Int) {
205 var childIndex = index
206 let child = nodes[childIndex]
207 var parentIndex = self.parentIndex(ofIndex: childIndex)
208
209 while childIndex > 0 && orderCriteria(child, nodes[parentIndex]) {
210 nodes[childIndex] = nodes[parentIndex]
211 childIndex = parentIndex
212 parentIndex = self.parentIndex(ofIndex: childIndex)
213 }
214
215 nodes[childIndex] = child
216 }
217
218 /**
219 * Looks at a parent node and makes sure it is still larger (max-heap) or
220 * smaller (min-heap) than its childeren.
221 */
222 internal mutating func shiftDown(from index: Int, until endIndex: Int) {
223 let leftChildIndex = self.leftChildIndex(ofIndex: index)
224 let rightChildIndex = leftChildIndex + 1
225
226 // Figure out which comes first if we order them by the sort function:
227 // the parent, the left child, or the right child. If the parent comes
228 // first, we‘re done. If not, that element is out-of-place and we make
229 // it "float down" the tree until the heap property is restored.
230 var first = index
231 if leftChildIndex < endIndex && orderCriteria(nodes[leftChildIndex], nodes[first]) {
232 first = leftChildIndex
233 }
234 if rightChildIndex < endIndex && orderCriteria(nodes[rightChildIndex], nodes[first]) {
235 first = rightChildIndex
236 }
237 if first == index { return }
238
239 nodes.swapAt(index, first)
240 shiftDown(from: first, until: endIndex)
241 }
242
243 internal mutating func shiftDown(_ index: Int) {
244 shiftDown(from: index, until: nodes.count)
245 }
246
247 }
248
249 // MARK: - Searching
250 extension Heap where T: Equatable {
251
252 /** Get the index of a node in the heap. Performance: O(n). */
253 public func index(of node: T) -> Int? {
254 return nodes.index(where: { $0 == node })
255 }
256
257 /** Removes the first occurrence of a node from the heap. Performance: O(n log n). */
258 @discardableResult public mutating func remove(node: T) -> T? {
259 if let index = index(of: node) {
260 return remove(at: index)
261 }
262 return nil
263 }
264 }
88ms
1 class Solution {
2
3 func findRedundantConnection(_ edges: [[Int]]) -> [Int] {
4 var N = 0
5 var graph = [Int: Set<Int>]()
6 for edge in edges {
7 graph[edge[0], default: []].insert(edge[1])
8 graph[edge[1], default: []].insert(edge[0])
9 N = max(N, edge[0], edge[1])
10 }
11 let source = edges[0][0]
12 for edge in edges.reversed() {
13 if isConnected(graph, edge, source, N) {
14 return edge
15 }
16 }
17 return edges.last!
18 }
19
20 func isConnected(_ graph: [Int: Set<Int>], _ edge: [Int], _ source: Int, _ N: Int) -> Bool {
21 var graph = graph
22 graph[edge[0]]!.remove(edge[1])
23 graph[edge[1]]!.remove(edge[0])
24 var stack = [Int]()
25 var visited = Set<Int>()
26 stack.append(source)
27 while !stack.isEmpty {
28 let node = stack.popLast()!
29 visited.insert(node)
30 for edge in graph[node] ?? [] {
31 if !visited.contains(edge) {
32 stack.append(edge)
33 }
34 }
35 }
36
37 return visited.count == N
38 }
39 }
112ms
1 class Solution {
2
3 let MAX_EDGE_VAL = 1000
4
5 func findRedundantConnection(_ edges: [[Int]]) -> [Int] {
6 var graph = [Int: [Int]]()
7
8 for edge in edges {
9 let u = edge[0]
10 let v = edge[1]
11 var visited = Set<Int>()
12 if hasPath(&graph, &visited, u, v) {
13 return [u, v]
14 }
15 graph[u] = graph[u] ?? [Int]()
16 graph[u]!.append(v)
17 graph[v] = graph[v] ?? [Int]()
18 graph[v]!.append(u)
19 }
20 return [-1, -1]
21 }
22
23 public func hasPath(_ graph: inout [Int: [Int]], _ visited: inout Set<Int>, _ source: Int, _ target: Int) -> Bool {
24 if source == target {
25 return true
26 }
27 if !graph.keys.contains(source) || !graph.keys.contains(target) {
28 return false
29 }
30 visited.insert(source)
31 if let neighbers = graph[source] {
32 for neighber in neighbers {
33 if visited.contains(neighber) {
34 continue
35 }
36 if hasPath(&graph, &visited, neighber, target) {
37 return true
38 }
39 }
40 }
41 return false
42 }
43 }
原文地址：https://www.cnblogs.com/strengthen/p/10500242.html
时间： 2024-10-10 07:46:32