leetcode200.岛屿数量

title: leetcode200.岛屿数量

题目描述：

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例1：

输入：grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出：1

示例2：

输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] 的值为 '0' 或 '1'

代码如下：

int colSize;
int size;

void inject(char **grid, int i, int j)
{
    if (i < 0 || i >=size || j < 0 || j >= colSize || grid[i][j] != '1') {
        return;
    }

    grid[i][j] = '2';
    inject(grid, i - 1, j);
    inject(grid, i + 1, j);
    inject(grid, i, j - 1);
    inject(grid, i, j + 1);
}

int numIslands(char** grid, int gridSize, int* gridColSize){
    int cntLand = 0;
    colSize = gridColSize[0];
    size = gridSize;
    for (int i = 0; i < size; ++i)
    {
        for (int j = 0; j < colSize; ++j)
        {
            if (grid[i][j] == '1') {
                inject(grid, i, j);
                ++cntLand;
            }
        }
    }
    return cntLand;
}