【手把手带你刷LeetCode】——10.岛屿数量（DFS）-CFANZ编程社区

【手把手带你刷LeetCode】——10.岛屿数量（DFS）

【前言】：

今天是力扣打卡第10天！

每天进步一点点，加油啦。

【手把手带你刷LeetCode】——10.岛屿数量（DFS）_职场和发展

原题：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

题解：

核心思想就是将‘1’的上下左右的‘1’同化成‘0’，如果不明白，先去体会代码哦

代码执行：

int numIslands(char** grid, int gridSize, int* gridColSize){
    //找递归边界
    if(grid == NULL || gridSize == 0)
    {
        return 0;
    }
    int row = gridSize;//行数
    int col = *gridColSize;//列数
    int count = 0;//用于计数
    int i = 0;
    int j = 0;
    //遍历这个二维网格
    for(i = 0; i < row; i++)
    {
        for(j = 0; j < col; j++)
        {
            if(grid[i][j] == '1')
            {
                count++;
                //将‘1’周围的‘1’全部同化成0
                dfs(grid, i, j, row, col);
            }
            
        }
    }
    return count;
}

void dfs(char** grid, int x, int y, int row, int col)
{
    //判断特殊情况
    if(x < 0 || x >= row || y < 0 || y >= col || grid[x][y] == '0')//注意哦，下标等于行数列数时也是不可以的哟，还要加上grid[x][y]是否为0这个判断条件
    {
        return;
    }
    grid[x][y] = '0';
    dfs(grid, x - 1, y, row, col);
    dfs(grid, x + 1, y, row, col);
    dfs(grid, x, y - 1, row, col);
    dfs(grid, x, y + 1, row, col);
}