[Swift]LeetCode934. 最短的桥 | Shortest Bridge
2024-06-06 07:06:44
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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/9903939.html
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In a given 2D binary array A, there are two islands. (An island is a 4-directionally connected group of 1s not connected to any other 1s.)
Now, we may change 0s to 1s so as to connect the two islands together to form 1 island.
Return the smallest number of 0s that must be flipped. (It is guaranteed that the answer is at least 1.)
Example 1:
Input: [[0,1],[1,0]] Output: 1
Example 2:
Input: [[0,1,0],[0,0,0],[0,0,1]] Output: 2
Example 3:
Input: [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]] Output: 1
Note:
1 <= A.length = A[0].length <= 100A[i][j] == 0orA[i][j] == 1
在给定的二维二进制数组 A 中，存在两座岛。（岛是由四面相连的 1 形成的一个最大组。）
现在，我们可以将 0 变为 1，以使两座岛连接起来，变成一座岛。
返回必须翻转的 0 的最小数目。（可以保证答案至少是 1。）
示例 1：
输入：[[0,1],[1,0]] 输出：1
示例 2：
输入：[[0,1,0],[0,0,0],[0,0,1]] 输出：2
示例 3：
输入：[[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]] 输出：1
提示：
1 <= A.length = A[0].length <= 100A[i][j] == 0或A[i][j] == 1
404ms
1 class Solution {
2 func shortestBridge(_ A: [[Int]]) -> Int {
3 let rowCount = A.count
4 if rowCount == 0 {
5 return 0
6 }
7
8 let colCount = A[0].count
9
10 var A = A
11
12 let coordinates = findFirstIsland(A: A)
13 let r1 = coordinates[0]
14 let c1 = coordinates[1]
15
16 var nextRiver = [Int]()
17 removeFirstIsland(A: &A, r: r1, c: c1, river: &nextRiver)
18
19 var distance = 0
20 while nextRiver.count > 0 {
21 let river = nextRiver
22 nextRiver = [Int]()
23
24 for coordinates in river {
25 let r = coordinates / rowCount
26 let c = coordinates % rowCount
27
28 let cell = A[r][c]
29 if cell == 0 {
30 A[r][c] = -1
31
32 if r > 0 {
33 // top
34 nextRiver.append((r - 1) * rowCount + c)
35 }
36 if (r + 1) < rowCount {
37 // bottom
38 nextRiver.append((r + 1) * rowCount + c)
39 }
40 if c > 0 {
41 nextRiver.append(r * rowCount + c - 1)
42 }
43 if (c + 1) < colCount {
44 nextRiver.append(r * rowCount + c + 1)
45 }
46 } else if cell == 1 {
47 return distance
48 }
49 }
50 distance += 1
51 }
52 return distance
53 }
54
55 func removeFirstIsland(A: inout [[Int]], r: Int, c: Int, river: inout [Int]) {
56 let rowCount = A.count
57 let colCount = A[0].count
58
59 let cell = A[r][c]
60 if cell == 0 {
61 // A[r][c] = 2
62 river.append(r * rowCount + c)
63 return
64 } else if cell == -1 {
65 return
66 }
67 A[r][c] = -1
68
69 if r > 0 {
70 // top
71 removeFirstIsland(A: &A, r: r - 1, c: c, river: &river)
72 }
73 if (r + 1) < rowCount {
74 // bottom
75 removeFirstIsland(A: &A, r: r + 1, c: c, river: &river)
76 }
77 if c > 0 {
78 removeFirstIsland(A: &A, r: r, c: c - 1, river: &river)
79 }
80 if (c + 1) < colCount {
81 removeFirstIsland(A: &A, r: r, c: c + 1, river: &river)
82 }
83 }
84
85 func findFirstIsland(A: [[Int]]) -> [Int] {
86 let rowCount = A.count
87 let colCount = A[0].count
88 var coordinates = Array(repeating: 0, count: 2)
89 var r = 0
90 var c = 0
91 while r < rowCount {
92 c = 0
93 while c < colCount {
94 if A[r][c] == 1 {
95 coordinates[0] = r
96 coordinates[1] = c
97 return coordinates
98 }
99 c += 1
100 }
101 r += 1
102 }
103 return coordinates
104 }
105 }428ms
1 class Solution {
2 func shortestBridge(_ A: [[Int]]) -> Int {
3 var a = A
4 var q = [(Int, Int)]()
5
6 loop: for i in 0 ..< A.count {
7 for j in 0 ..< A[i].count {
8 if A[i][j] == 1 {
9 dfs(&a, i, j, &q)
10 break loop
11 }
12 }
13 }
14
15 let dir = [0 , 1 , 0 , -1 , 0]
16 var step = 0
17 while !q.isEmpty {
18
19 var size = q.count
20 while size > 0 {
21 let (i, j) = q.removeFirst()
22 for index in 0 ..< 4 {
23 let ti = i + dir[index]
24 let tj = j + dir[index + 1]
25 if ti < 0 || ti >= a.count || tj < 0 || tj >= a[i].count || a[ti][tj] == 2 {
26 continue
27 }
28 if a[ti][tj] == 1 {
29 return step
30 } else {
31 a[ti][tj] = 2
32 q.append((ti,tj))
33
34 }
35 }
36 size -= 1
37 }
38 step += 1
39 }
40
41 return -1
42 }
43
44 func dfs(_ A: inout [[Int]], _ i: Int, _ j: Int, _ q: inout [(Int, Int)]) {
45 if i < 0 || i >= A.count || j < 0 || j >= A[i].count || A[i][j] != 1 {
46 return
47 } else {
48 A[i][j] = 2
49 q.append((i, j))
50 dfs(&A, i - 1, j, &q)
51 dfs(&A, i + 1, j, &q)
52 dfs(&A, i, j - 1, &q)
53 dfs(&A, i, j + 1, &q)
54 }
55 }
56 }476ms
1 class Solution {
2 func shortestBridge(_ A: [[Int]]) -> Int {
3 let R = A.count
4 let C = A[0].count
5 var qs = [(Int, Int)]()
6 var color = 1
7 var v = [[Int]](repeating: [Int](repeating: 0, count: C), count: R)
8 var ans = 0
9 func next(_ r: Int, _ c: Int, _ d: inout [String: (Int, Int)]) -> Bool {
10 var ns = [(r+1, c), (r-1, c), (r, c+1), (r, c-1)]
11 for n in ns {
12 guard 0 ..< R ~= n.0, 0 ..< C ~= n.1 else {
13 continue
14 }
15 guard v[n.0][n.1] == 0 else {
16 if v[n.0][n.1] != v[r][c] {
17 //print(n, r, c)
18 return true
19 }
20 continue
21 }
22 v[n.0][n.1] = v[r][c]
23 d["\(n.0), \(n.1)"] = (n.0, n.1)
24 }
25 return false
26 }
27 func paint(_ r: Int, _ c: Int) {
28 guard v[r][c] == 0 else {
29 return
30 }
31 v[r][c] = color
32 var qs = [(r, c)]
33 while !qs.isEmpty {
34 for i in 0 ..< qs.count {
35 let (r, c) = qs.remove(at: 0)
36 let ns = [(r+1, c), (r-1, c), (r, c+1), (r, c-1)]
37 for n in ns {
38 guard 0 ..< R ~= n.0, 0 ..< C ~= n.1 else {
39 continue
40 }
41 guard v[n.0][n.1] == 0, A[n.0][n.1] == 1 else {
42 continue
43 }
44 v[n.0][n.1] = color
45 qs.append(n)
46 }
47 }
48 }
49 color += 1
50 }
51 for r in 0 ..< R {
52 for c in 0 ..< C {
53 if A[r][c] == 1 {
54 paint(r, c)
55 if v[r][c] == 1 {
56 qs.append((r, c))
57 }
58 }
59 }
60 }
61 //print(v)
62 while !qs.isEmpty {
63 //print(qs)
64 var d = [String: (Int, Int)]()
65 for i in 0 ..< qs.count {
66 let (r, c) = qs.remove(at: 0)
67 guard next(r, c, &d) == false else {
68 //print(v)
69 return ans
70 }
71 }
72 ans += 1
73 qs = Array(d.values)
74 }
75
76 return ans
77 }
78 }484ms
1 class Solution {
2 struct Coor {
3 var x:Int;
4 var y:Int;
5 }
6 var visited:[[Bool]] = [[Bool]]();
7 var grids:[[Int]]!;
8 func shortestBridge(_ A: [[Int]]) -> Int {
9 var step:Int = -1;
10 grids = A;
11 for i in 0 ..< A.count{
12 let oneRow = Array<Bool>(repeating: false, count: A[i].count);
13 visited.append(oneRow);
14 }
15 var i = 0;
16 var j = 0;
17 for ii in 0 ..< A.count{
18 for jj in 0 ..< A[ii].count{
19 if (A[ii][jj] == 1){
20 i = ii;
21 j = jj;
22 break;
23 }
24 }
25 }
26 let island_A = completeThisIsland(x: i, y: j);
27 // resetVisited()
28 var queue = Array<Coor>();
29 var steps = Array<Int>(repeating: 0, count: island_A.count);
30 queue.append(contentsOf: island_A);
31 var found = false;
32 while (!found){
33 let co = queue.removeFirst();
34 let thisStep = steps.removeFirst();
35 let neighbors = [Coor(x: co.x-1, y: co.y),Coor(x: co.x+1, y: co.y),Coor(x: co.x, y: co.y+1),Coor(x:co.x,y:co.y-1)];
36 for neighbor in neighbors{
37 if valid(coor: neighbor) && !visited[neighbor.x][neighbor.y]{
38 if grids[neighbor.x][neighbor.y] == 1{
39 step = thisStep;
40 found = true;
41 break;
42 }else{
43 queue.append(neighbor)
44 steps.append(thisStep+1)
45 visited[neighbor.x][neighbor.y] = true;
46 }
47 // queue.append(neighbor);
48 // visited[neighbor.x][neighbor.y] = true;
49 }
50 }
51 }
52
53 return step;
54 }
55 func resetVisited() -> Void {
56 for var i in visited{
57 for j in 0 ..< i.count{
58 i[j] = false;
59 }
60 }
61 }
62 func valid(coor:Coor) -> Bool {
63 return coor.x > -1 && coor.x < grids.count && coor.y > -1 && coor.y < grids[coor.x].count;
64 }
65 func completeThisIsland(x:Int,y:Int) -> [Coor] {
66 var result = [Coor]();
67 var queue = [Coor]();
68 queue.append(Coor(x: x, y: y))
69 visited[x][y] = true;
70 while(!queue.isEmpty){
71 let co = queue.removeFirst();
72 // visited[co.x][co.y] = true;
73 result.append(co);
74 let neighbors = [Coor(x: co.x-1, y: co.y),Coor(x: co.x+1, y: co.y),Coor(x: co.x, y: co.y+1),Coor(x:co.x,y:co.y-1)];
75 for neighbor in neighbors{
76 if valid(coor: neighbor) && grids[neighbor.x][neighbor.y] == 1 && !visited[neighbor.x][neighbor.y]{
77 queue.append(neighbor);
78 visited[neighbor.x][neighbor.y] = true;
79 }
80 }
81 }
82 return result;
83 }
84 }624ms
1 class Solution {
2 func shortestBridge(_ A: [[Int]]) -> Int {
3 var a = A
4 //print(A)
5 var edI = -1
6 var edJ = -1
7 for i in (0..<A.count) {
8 for j in (0..<A[i].count) {
9 if A[i][j] == 1 {
10 edI = i
11 edJ = j
12 break
13 }
14 }
15 if edI != -1 { break }
16 }
17
18 //print(edI, edJ)
19 bfs(&a, edI, edJ)
20
21 for i in (0..<a.count) {
22 for j in (0..<a[i].count) {
23 if a[i][j] == 1 {
24 a[i][j] = -2
25 }
26 }
27 }
28 //print(a)
29
30 return dfs(&a, edI, edJ) - 1
31 }
32
33 func bfs(_ A: inout [[Int]], _ i: Int, _ j: Int) {
34 A[i][j] = -1
35 if i+1 < A.count && A[i+1][j] == 1 {
36 A[i+1][j] = -1
37 bfs(&A, i+1, j)
38 }
39
40 if j + 1 < A[i].count && A[i][j+1] == 1 {
41 A[i][j+1] = -1
42 bfs(&A, i, j+1)
43 }
44
45 if i > 0 && A[i-1][j] == 1 {
46 A[i-1][j] = -1
47 bfs(&A, i-1, j)
48 }
49
50 if j > 0 && A[i][j-1] == 1 {
51 A[i][j-1] = -1
52 bfs(&A, i, j-1)
53 }
54 }
55
56 func dfs(_ A: inout [[Int]], _ i: Int, _ j: Int) -> Int {
57 var queue: [Int] = []
58 queue.append(i * 200 + j)
59 while queue.count > 0 {
60 //print(queue)
61 //print(A)
62 let curI = queue[0] / 200
63 let curJ = queue[0] % 200
64 if A[curI][curJ] == -1 { A[curI][curJ] = -3 }
65 queue.remove(at: 0)
66 if curI+1 < A.count {
67 if A[curI+1][curJ] == -2 {
68 return A[curI][curJ] + 1
69 } else if A[curI+1][curJ] == -1 {
70 queue.insert((curI+1)*200+curJ, at: 0)
71 } else if A[curI+1][curJ] == 0 {
72 if A[curI][curJ] < 0 {
73 A[curI+1][curJ] = 1
74 } else {
75 A[curI+1][curJ] = A[curI][curJ] + 1
76 }
77 queue.append((curI+1)*200+curJ)
78 }
79 }
80
81 if curJ+1 < A[curI].count {
82 if A[curI][curJ+1] == -2 {
83 return A[curI][curJ] + 1
84 } else if A[curI][curJ+1] == -1 {
85 queue.insert((curI)*200+curJ+1, at: 0)
86 } else if A[curI][curJ+1] == 0 {
87 if A[curI][curJ] < 0 {
88 A[curI][curJ+1] = 1
89 } else {
90 A[curI][curJ+1] = A[curI][curJ] + 1
91 }
92 queue.append((curI)*200+curJ+1)
93 }
94 }
95
96 if curI > 0 {
97 if A[curI-1][curJ] == -2 {
98 return A[curI][curJ] + 1
99 } else if A[curI-1][curJ] == -1 {
100 queue.insert((curI-1)*200+curJ, at: 0)
101 } else if A[curI-1][curJ] == 0 {
102 if A[curI][curJ] < 0 {
103 A[curI-1][curJ] = 1
104 } else {
105 A[curI-1][curJ] = A[curI][curJ] + 1
106 }
107 queue.append((curI-1)*200+curJ)
108 }
109 }
110
111 if curJ > 0 {
112 if A[curI][curJ-1] == -2 {
113 return A[curI][curJ] + 1
114 } else if A[curI][curJ-1] == -1 {
115 queue.insert((curI)*200+curJ-1, at: 0)
116 } else if A[curI][curJ-1] == 0 {
117 if A[curI][curJ] < 0 {
118 A[curI][curJ-1] = 1
119 } else {
120 A[curI][curJ-1] = A[curI][curJ] + 1
121 }
122 queue.append((curI)*200+curJ-1)
123 }
124 }
125
126 }
127 return 0
128 }
129 }648ms
1 class Solution {
2 func shortestBridge(_ A: [[Int]]) -> Int {
3 var n:Int = A.count, m = A[0].count
4 var can:[[Bool]] = [[Bool]](repeating: [Bool](repeating: false, count: m), count: n)
5 var dr:[Int] = [1, 0, -1, 0 ]
6 var dc:[Int] = [0, 1, 0, -1 ]
7 var d:[[Int]] = [[Int]](repeating: [Int](repeating: 99999999, count: m), count: n)
8
9 var gq:Queue<[Int]> = Queue<[Int]>()
10 inner:
11 for i in 0..<n
12 {
13 for j in 0..<m
14 {
15 if A[i][j] == 1
16 {
17 var q:Queue<[Int]> = Queue<[Int]>()
18 q.enQueue([i,j])
19 can[i][j] = true
20 while(!q.isEmpty())
21 {
22 var cur:[Int] = q.deQueue()!
23 gq.enQueue(cur)
24 var r:Int = cur[0], c:Int = cur[1]
25 d[r][c] = 0
26 for k in 0..<4
27 {
28 var nr:Int = r + dr[k], nc:Int = c + dc[k]
29 if nr >= 0 && nr < n && nc >= 0 && nc < m && A[nr][nc] == 1 && !can[nr][nc]
30 {
31 can[nr][nc] = true
32 q.enQueue([nr, nc])
33 }
34 }
35 }
36 break inner
37 }
38 }
39 }
40
41 while(!gq.isEmpty())
42 {
43 var cur:[Int] = gq.deQueue()!
44 var r:Int = cur[0], c:Int = cur[1]
45 for k in 0..<4
46 {
47 var nr:Int = r + dr[k], nc:Int = c + dc[k]
48 if nr >= 0 && nr < n && nc >= 0 && nc < m && d[nr][nc] > d[r][c] + 1
49 {
50 d[nr][nc] = d[r][c] + 1
51 gq.enQueue([nr, nc])
52 }
53 }
54 }
55 var ret:Int = 9999999
56 for i in 0..<n
57 {
58 for j in 0..<m
59 {
60 if !can[i][j] && A[i][j] == 1
61 {
62 ret = min(ret, d[i][j])
63 }
64 }
65 }
66 return ret-1
67 }
68 }
69
70 public struct Queue<T> {
71
72 // 泛型数组：用于存储数据元素
73 fileprivate var queue: [T]
74
75 // 返回队列中元素的个数
76 public var count: Int {
77 return queue.count
78 }
79
80 // 构造函数：创建一个空的队列
81 public init() {
82 queue = [T]()
83 }
84
85 //通过既定数组构造队列
86 init(_ arr:[T]){
87 queue = arr
88 }
89
90 // 如果定义了默认值，则可以在调用函数时省略该参数
91 init(_ elements: T...) {
92 queue = elements
93 }
94
95 // 检查队列是否为空
96 // - returns: 如果队列为空，则返回true，否则返回false
97 public func isEmpty() -> Bool {
98 return queue.isEmpty
99 }
100
101 // 入队列操作：将元素添加到队列的末尾
102 public mutating func enQueue(_ element: T) {
103 queue.append(element)
104 }
105
106 // 出队列操作：删除并返回队列中的第一个元素
107 public mutating func deQueue() -> T? {
108 return queue.removeFirst()
109 }
110 }转载于:https://www.cnblogs.com/strengthen/p/9903939.html
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