[LeetCode 200] Number of Islands

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本文介绍了一种使用深度优先搜索(DFS)加剪枝的方法来计算二维网格中岛屿的数量。通过遍历每个单元格并标记已访问过的陆地，有效地解决了岛屿计数问题。

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:

Answer: 1

Example 2:

Answer: 3

solution:

DFS + pruning

found 1 then search its up, left, down, right. Meanwhile change value to 2, pruning.

public int numIslands(char[][] grid) {
        int res = 0;
        int row = grid.length;
        if(row <=0) return 0;
        int col = grid[0].length;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(grid[i][j] == '1'){
                    res++;
                    populateAdj(grid, i, j);
                }
            }
        }
        return res;
    }
    public void populateAdj(char[][] grid, int x, int y){
        if(x>= grid.length || y>=grid[0].length || x<0 || y<0) return;
        if(grid[x][y] == '1') {
            grid[x][y] = '2';
            populateAdj(grid, x-1, y);
            populateAdj(grid, x+1, y);
            populateAdj(grid, x, y-1);
            populateAdj(grid, x, y+1);
        }
    }