[Swift]LeetCode684. 冗余连接 | Redundant Connection
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➤微信公众号：山青咏芝（shanqingyongzhi）
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➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址： https://www.cnblogs.com/strengthen/p/10500242.html
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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 public struct Heap<T> {
46
47 /** The array that stores the heap's nodes. */
48 var nodes = [T]()
49
50 /**
51 * Determines how to compare two nodes in the heap.
52 * Use '>' for a max-heap or '<' for a min-heap,
53 * or provide a comparing method if the heap is made
54 * of custom elements, for example tuples.
55 */
56 private var orderCriteria: (T, T) -> Bool
57
58 /**
59 * Creates an empty heap.
60 * The sort function determines whether this is a min-heap or max-heap.
61 * For comparable data types, > makes a max-heap, < makes a min-heap.
62 */
63 public init(sort: @escaping (T, T) -> Bool) {
64 self.orderCriteria = sort
65 }
66
67 /**
68 * Creates a heap from an array. The order of the array does not matter;
69 * the elements are inserted into the heap in the order determined by the
70 * sort function. For comparable data types, '>' makes a max-heap,
71 * '<' makes a min-heap.
72 */
73 public init(array: [T], sort: @escaping (T, T) -> Bool) {
74 self.orderCriteria = sort
75 configureHeap(from: array)
76 }
77
78 /**
79 * Configures the max-heap or min-heap from an array, in a bottom-up manner.
80 * Performance: This runs pretty much in O(n).
81 */
82 private mutating func configureHeap(from array: [T]) {
83 nodes = array
84 for i in stride(from: (nodes.count/2-1), through: 0, by: -1) {
85 shiftDown(i)
86 }
87 }
88
89 public var isEmpty: Bool {
90 return nodes.isEmpty
91 }
92
93 public var count: Int {
94 return nodes.count
95 }
96
97 /**
98 * Returns the index of the parent of the element at index i.
99 * The element at index 0 is the root of the tree and has no parent.
100 */
101 @inline(__always) internal func parentIndex(ofIndex i: Int) -> Int {
102 return (i - 1) / 2
103 }
104
105 /**
106 * Returns the index of the left child of the element at index i.
107 * Note that this index can be greater than the heap size, in which case
108 * there is no left child.
109 */
110 @inline(__always) internal func leftChildIndex(ofIndex i: Int) -> Int {
111 return 2*i + 1
112 }
113
114 /**
115 * Returns the index of the right child of the element at index i.
116 * Note that this index can be greater than the heap size, in which case
117 * there is no right child.
118 */
119 @inline(__always) internal func rightChildIndex(ofIndex i: Int) -> Int {
120 return 2*i + 2
121 }
122
123 /**
124 * Returns the maximum value in the heap (for a max-heap) or the minimum
125 * value (for a min-heap).
126 */
127 public func peek() -> T? {
128 return nodes.first
129 }
130
131 /**
132 * Adds a new value to the heap. This reorders the heap so that the max-heap
133 * or min-heap property still holds. Performance: O(log n).
134 */
135 public mutating func insert(_ value: T) {
136 nodes.append(value)
137 shiftUp(nodes.count - 1)
138 }
139
140 /**
141 * Adds a sequence of values to the heap. This reorders the heap so that
142 * the max-heap or min-heap property still holds. Performance: O(log n).
143 */
144 public mutating func insert<S: Sequence>(_ sequence: S) where S.Iterator.Element == T {
145 for value in sequence {
146 insert(value)
147 }
148 }
149
150 /**
151 * Allows you to change an element. This reorders the heap so that
152 * the max-heap or min-heap property still holds.
153 */
154 public mutating func replace(index i: Int, value: T) {
155 guard i < nodes.count else { return }
156
157 remove(at: i)
158 insert(value)
159 }
160
161 /**
162 * Removes the root node from the heap. For a max-heap, this is the maximum
163 * value; for a min-heap it is the minimum value. Performance: O(log n).
164 */
165 @discardableResult public mutating func remove() -> T? {
166 guard !nodes.isEmpty else { return nil }
167
168 if nodes.count == 1 {
169 return nodes.removeLast()
170 } else {
171 // Use the last node to replace the first one, then fix the heap by
172 // shifting this new first node into its proper position.
173 let value = nodes[0]
174 nodes[0] = nodes.removeLast()
175 shiftDown(0)
176 return value
177 }
178 }
179
180 /**
181 * Removes an arbitrary node from the heap. Performance: O(log n).
182 * Note that you need to know the node's index.
183 */
184 @discardableResult public mutating func remove(at index: Int) -> T? {
185 guard index < nodes.count else { return nil }
186
187 let size = nodes.count - 1
188 if index != size {
189 nodes.swapAt(index, size)
190 shiftDown(from: index, until: size)
191 shiftUp(index)
192 }
193 return nodes.removeLast()
194 }
195
196 /**
197 * Takes a child node and looks at its parents; if a parent is not larger
198 * (max-heap) or not smaller (min-heap) than the child, we exchange them.
199 */
200 internal mutating func shiftUp(_ index: Int) {
201 var childIndex = index
202 let child = nodes[childIndex]
203 var parentIndex = self.parentIndex(ofIndex: childIndex)
204
205 while childIndex > 0 && orderCriteria(child, nodes[parentIndex]) {
206 nodes[childIndex] = nodes[parentIndex]
207 childIndex = parentIndex
208 parentIndex = self.parentIndex(ofIndex: childIndex)
209 }
210
211 nodes[childIndex] = child
212 }
213
214 /**
215 * Looks at a parent node and makes sure it is still larger (max-heap) or
216 * smaller (min-heap) than its childeren.
217 */
218 internal mutating func shiftDown(from index: Int, until endIndex: Int) {
219 let leftChildIndex = self.leftChildIndex(ofIndex: index)
220 let rightChildIndex = leftChildIndex + 1
221
222 // Figure out which comes first if we order them by the sort function:
223 // the parent, the left child, or the right child. If the parent comes
224 // first, we're done. If not, that element is out-of-place and we make
225 // it "float down" the tree until the heap property is restored.
226 var first = index
227 if leftChildIndex < endIndex && orderCriteria(nodes[leftChildIndex], nodes[first]) {
228 first = leftChildIndex
229 }
230 if rightChildIndex < endIndex && orderCriteria(nodes[rightChildIndex], nodes[first]) {
231 first = rightChildIndex
232 }
233 if first == index { return }
234
235 nodes.swapAt(index, first)
236 shiftDown(from: first, until: endIndex)
237 }
238
239 internal mutating func shiftDown(_ index: Int) {
240 shiftDown(from: index, until: nodes.count)
241 }
242
243 }
244
245 // MARK: - Searching
246 extension Heap where T: Equatable {
247
248 /** Get the index of a node in the heap. Performance: O(n). */
249 public func index(of node: T) -> Int? {
250 return nodes.index(where: { $0 == node })
251 }
252
253 /** Removes the first occurrence of a node from the heap. Performance: O(n log n). */
254 @discardableResult public mutating func remove(node: T) -> T? {
255 if let index = index(of: node) {
256 return remove(at: index)
257 }
258 return nil
259 }
260 }
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 }
