leetcode 200. Number of Islands(岛的数量-连通区域数）

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:

Input:
11110
11010
11000
00000

Output: 1

把1看作小岛，0看作水，而且数组的外围默认是水。
上下左右算连通，斜方向不算。
问小岛有多少个

思路：

DFS
本质上是求1的连通区域，
因此每走到一个1处，搜索它上下左右是否为1，是1就算同一片区域，同时搜索完后将这个位置的1改为0，防止重复搜索。
如果不想修改原数组，可用和grid同样大小的boolean数组表示每个位置是否已经被访问过。

    public int numIslands(char[][] grid) {
        int num = 0;
        int rows = grid.length;
        int cols = grid[0].length;
        
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                num += grid[i][j] - '0';
                dfs(grid, i, j, rows, cols);
            }
        }
        //System.out.println(grid[0][0]);
        return num;
    }
    
    public void dfs(char[][] grid, int i, int j, int rows, int cols) {
        if (i < 0 || i >= rows || j < 0 || j >= cols || grid[i][j] == '0') {
            return;
        }
        
        grid[i][j] = '0';
        
        dfs(grid, i-1, j, rows, cols); //up
        dfs(grid, i+1, j, rows, cols); //down
        dfs(grid, i, j-1, rows, cols); //left
        dfs(grid, i, j+1, rows, cols); //right
    }

注释的地方为表示访问过的visited数组

class Solution {
    //boolean[][] visited;
    int m;
    int n;
    public int numIslands(char[][] grid) {
        int num = 0;
        m = grid.length;
        n = grid[0].length;
        //visited = new boolean[m][n];
        
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == '0') continue;
                dfs(grid, i, j);
                num ++;
            }
        }
        return num;
    }
    
    void dfs(char[][] grid, int i, int j) {
        if(i < 0 || i >=m || j < 0 || j >=n) return;
        if(grid[i][j] == '0') return;
        
        grid[i][j] = '0';
        //visited[i][j] = true;
        dfs(grid, i-1, j);
        dfs(grid, i, j-1);
        dfs(grid, i, j+1);
        dfs(grid, i+1, j);
    }
    
    
}