200. Number of Islands

Posted 2020-10-01

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

bfs : 不使用visited 数组 则将其

思路很简单,遍历矩阵数联通的1块, 结果增, 防止再次遍历矩阵时重复将数过的1标记或转化为‘0‘ 或 ‘2‘

 

public class Solution {
    int[][] dirs = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
        
        int m = grid.length;
        int n = grid[0].length;
        //boolean[][] visited = new boolean[m][n];
        int ans = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] != ‘1‘) {
                    continue;
                }
                // true;
                grid[i][j] = ‘2‘;
                bfs(grid, new int[]{i, j});
                ans++;
            }
        }
        return ans;
        
    }
    private void bfs(char[][] grid,  int[] cell) {
        int m = grid.length;
        int n = grid[0].length;
        Queue<int[]> q = new LinkedList<>();
        q.offer(cell);
        while (!q.isEmpty()) {
            int[] cur = q.poll();
            for (int[] dir : dirs) {
                int x = dir[0] + cur[0];
                int y = dir[1] + cur[1];
                if (x >= m || x < 0 || y >= n || y < 0 || grid[x][y] != ‘1‘) {
                    continue;
                }
                grid[x][y] = ‘2‘;
                q.offer(new int[]{x, y});
            }
        }
    }
}

dfs

把入队改成自调用就ok了

int[][] dirs = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
        
        int m = grid.length;
        int n = grid[0].length;
        //boolean[][] visited = new boolean[m][n];
        int ans = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] != ‘1‘) {
                    continue;
                }
                // true;
                grid[i][j] = ‘0‘;
                bfs(grid, new int[]{i, j});
                ans++;
            }
        }
        return ans;
        
    }
    private void bfs(char[][] grid,  int[] cell) {
        int m = grid.length;
        int n = grid[0].length;
        
            for (int[] dir : dirs) {
                int x = dir[0] + cell[0];
                int y = dir[1] + cell[1];
                if (x >= m || x < 0 || y >= n || y < 0 || grid[x][y] != ‘1‘) {
                    continue;
                }
                grid[x][y] = ‘0‘;
                dfs(grid, new int[]{x, y});
            }
        
    }

 

和529. Minesweeper不同是因为529只是根据一个输入值找到以此的全部路径, 而此时是矩阵所有的点都可能为起点