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  • [LeetCode] 200. Number of Islands 岛屿的数量

    Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

    Example 1:

    11110
    11010
    11000
    00000

    Answer: 1

    Example 2:

    11000
    11000
    00100
    00011

    Answer: 3

    Credits:
    Special thanks to @mithmatt for adding this problem and creating all test cases.

     本质是求矩阵中连续区域的个数, 可以用BFS, DFS, 或者 Union Find来解。

    Java: BFS

    class Coordinate {
        int x, y;
        public Coordinate(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
    
    public class Solution {
        public int numIslands(boolean[][] grid) {
            if (grid == null || grid.length == 0 || grid[0].length == 0) {
                return 0;
            }
            
            int n = grid.length;
            int m = grid[0].length;
            int islands = 0;
            
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (grid[i][j]) {
                        markByBFS(grid, i, j);
                        islands++;
                    }
                }
            }
            
            return islands;
        }
        
        private void markByBFS(boolean[][] grid, int x, int y) {
            int[] directionX = {0, 1, -1, 0};
            int[] directionY = {1, 0, 0, -1};       
            Queue<Coordinate> queue = new LinkedList<>();      
            queue.offer(new Coordinate(x, y));
            grid[x][y] = false;
            
            while (!queue.isEmpty()) {
                Coordinate coor = queue.poll();
                for (int i = 0; i < 4; i++) {
                    Coordinate adj = new Coordinate(
                        coor.x + directionX[i],
                        coor.y + directionY[i]
                    );
                    if (!inBound(adj, grid)) {
                        continue;
                    }
                    if (grid[adj.x][adj.y]) {
                        grid[adj.x][adj.y] = false;
                        queue.offer(adj);
                    }
                }
            }
        }
        
        private boolean inBound(Coordinate coor, boolean[][] grid) {
            int n = grid.length;
            int m = grid[0].length;
            
            return coor.x >= 0 && coor.x < n && coor.y >= 0 && coor.y < m;
        }
    }
    

    Java: DFS

    public class Solution {
        private int m, n;
        public void dfs(boolean[][] grid, int i, int j) {
            if (i < 0 || i >= m || j < 0 || j >= n) return;
            
            if (grid[i][j]) {
                grid[i][j] = false;
                dfs(grid, i - 1, j);
                dfs(grid, i + 1, j);
                dfs(grid, i, j - 1);
                dfs(grid, i, j + 1);
            }
        }
    
        public int numIslands(boolean[][] grid) {
            m = grid.length;
            if (m == 0) return 0;
            n = grid[0].length;
            if (n == 0) return 0;
            
            int ans = 0;
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    if (!grid[i][j]) continue;
                    ans++;
                    dfs(grid, i, j);
                }
            }
            return ans;
        }
    }
    

    Java: Union Find

    class UnionFind { 
        private int[] father = null;
        private int count;
    
        private int find(int x) {
            if (father[x] == x) {
                return x;
            }
            return father[x] = find(father[x]);
        }
    
        public UnionFind(int n) {
            // initialize your data structure here.
            father = new int[n];
            for (int i = 0; i < n; ++i) {
                father[i] = i;
            }
        }
    
        public void connect(int a, int b) {
            int root_a = find(a);
            int root_b = find(b);
            if (root_a != root_b) {
                father[root_a] = root_b;
                count --;
            }
        }
            
        public int query() {
            return count;
        }
        
        public void set_count(int total) {
            count = total;
        }
    }
    
    public class Solution {
        public int numIslands(boolean[][] grid) {
            int count = 0;
            int n = grid.length;
            if (n == 0)
                return 0;
            int m = grid[0].length;
            if (m == 0)
                return 0;
            UnionFind union_find = new UnionFind(n * m);
            
            int total = 0;
            for(int i = 0;i < grid.length; ++i)
                for(int j = 0;j < grid[0].length; ++j)
                if (grid[i][j])
                    total ++;
        
            union_find.set_count(total);
            for(int i = 0;i < grid.length; ++i)
                for(int j = 0;j < grid[0].length; ++j)
                if (grid[i][j]) {
                    if (i > 0 && grid[i - 1][j]) {
                        union_find.connect(i * m + j, (i - 1) * m + j);
                    }
                    if (i <  n - 1 && grid[i + 1][j]) {
                        union_find.connect(i * m + j, (i + 1) * m + j);
                    }
                    if (j > 0 && grid[i][j - 1]) {
                        union_find.connect(i * m + j, i * m + j - 1);
                    }
                    if (j < m - 1 && grid[i][j + 1]) {
                        union_find.connect(i * m + j, i * m + j + 1);
                    }
                }
            return union_find.query();
        }
    }
    

    Python: BFS

    class Solution:
        def numIslands(self, grid):
            m = len(grid)
            if m == 0:
                return 0
            n = len(grid[0])
            visit = [[False for i in range(n)]for j in range(m)]
            def check(x, y):
                if x >= 0 and x < m and y >= 0 and y < n and grid[x][y] and visit[x][y] == False:
                    return True
            def bfs(x,y):
                nbrow = [1, 0, -1, 0]
                nbcol = [0, 1, 0, -1]
                q =[(x,y)]
                while len(q) > 0:
                    x = q[0][0]
                    y = q[0][1]
                    q.pop(0)
                    for k in range(4):
                        newx = x + nbrow[k]
                        newy = y + nbcol[k]
                        if check(newx, newy):
                            visit[newx][newy] = True
                            q.append((newx,newy))
                
            count = 0
            for row in range(m):
                for col in range(n):
                    if check(row,col):
                        visit[row][col] = True
                        bfs(row,col)
                        count+=1
            return count
    

    Python: DFS

    class Solution:
        def numIslands(self, grid):
            if not grid:
                return 0
        
            row = len(grid)
            col = len(grid[0])          
            count = 0
            for i in xrange(row):
                for j in xrange(col):
                    if grid[i][j] == '1':
                        self.dfs(grid, row, col, i, j)
                        count += 1
            return count
    
        def dfs(self, grid, row, col, x, y):
            if grid[x][y] == '0':
                return
            grid[x][y] = '0'
        
            if x != 0:
                self.dfs(grid, row, col, x - 1, y)
            if x != row - 1:
                self.dfs(grid, row, col, x + 1, y)
            if y != 0:
                self.dfs(grid, row, col, x, y - 1)
            if y != col - 1:
                self.dfs(grid, row, col, x, y + 1)
    

    Python: DFS

    class Solution(object):
        def findIslands(self, M):
           if not M:
               return 0
    
            res = 0
            for i in xrange(len(M)):
                for j in xrange(len(M[0])):
                    if M[i][j] == 1:
                        res += 1
                        self.dfs(M, i, j)
    
            return res
    
        def dfs(self, m, x, y):
            if m[x][y] == 1:
                m[x][y] = 0
                if x > 0:
                    self.dfs(m, x - 1, y)
                if y > 0:
                    self.dfs(m, x, y - 1)
                if x < len(m) - 1:
                    self.dfs(m, x + 1, y)
                if y < len(m[0]) - 1:
                    self.dfs(m, x, y + 1)  

    Python: BFS

    class Solution(object):
        def findIslands(self, M):
            if not M:
                return 0
    
            res = 0
            for i in xrange(len(M)):
                for j in xrange(len(M[0])):
                    if M[i][j] == 1:
                        res += 1
                        print res
                        self.bfs(M, i, j)
    
            return res
    
        def checkPoint(self, m, x, y):
            if x < 0 or y < 0 or x > len(m) - 1 or y > len(m[0]) - 1 or m[x][y] == 0:
                return False
    
            return True
    
        def bfs(self, m, x, y):
            x_row = [0, 0, -1, 1]
            y_col = [-1, 1, 0, 0]
            queue = [(x, y)]
            while len(queue) > 0:
                point = queue.pop(0)
                row = point[0]
                col = point[1]
                if m[row][col] == 1:
                    m[row][col] = 0
                    for i in xrange(4):
                        if self.checkPoint(m, row + x_row[i], col + y_col[i]):
                            queue.append((row + x_row[i], col + y_col[i])) 

    Python: DFS

    class Solution:
        def numIslands(self, grid):
            m = len(grid)
            if m == 0:
                return 0
            n = len(grid[0])
            visit = [[False for i in range(n)]for j in range(m)]
            def check(x, y):
                if x >= 0 and x < m and y >= 0 and y < n and grid[x][y] and visit[x][y] == False:
                    return True
            def dfs(x, y):
                nbrow = [1,0,-1,0]
                nbcol = [0,1,0,-1]
                for k in range(4):
                    newx = x + nbrow[k]
                    newy = y + nbcol[k]
                    if check(newx, newy):
                        visit[newx][newy] = True
                        dfs(newx,newy)
            count = 0
            for row in range(m):
                for col in range(n):
                    if check(row, col):
                        visit[row][col] = True
                        dfs(row, col)
                        count+=1
            return count
    

      

    类似题目:

    [LeetCode] 305. Number of Islands II 岛屿的数量 II

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    [LeetCode] 79. Word Search 单词搜索

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/8487025.html
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