leetcode 力扣 200 岛屿数量

深度优先遍历

算法思路

• 别想太多，只要是1就认为是岛屿，num_isLands++
• 从grid[i][j] == 1开始dfs，把这个岛变成海，完了

public int numIslands(char[][] grid) {
        int num_isLands = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    num_isLands++;
                    dfs(grid, i, j);
                }
            }
        }

        return num_isLands;
    }

    private void dfs(char[][] grid, int _row, int _column) {
        if (_row < 0 || _row >= grid.length 
            || _column < 0 || _column >= grid[0].length
            || grid[_row][_column] == '0') {
            return;
        }

        grid[_row][_column] = '0';

        dfs(grid, _row, _column + 1);
        dfs(grid, _row + 1, _column);
        dfs(grid, _row, _column - 1);
        dfs(grid, _row - 1, _column);
    }