200. 岛屿数量

200. 岛屿数量

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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'

题解：

bug:dfs函数中，if (grid[r][c] != 1 || r < 0 || r > row - 1 || c < 0 || c > column - 1) { return; }应该先判断越界再判断是否为1，先后顺序不能乱。修改后函数能正常输出了，原来是自己会自动退出，但是也不报错，很恶心，if()，while()，这里面的判断多个的顺序一定要注意区分。

#include <stdio.h>
#include <stdlib.h>

void dfs(int** grid, int r, int c, int row, int column)
{
    if (r < 0 || r > row - 1 || c < 0 || c > column - 1 || grid[r][c] != 1 )
    {
        return;
    }
    grid[r][c] = 2;
    dfs(grid, r + 1, c, row, column);
    dfs(grid, r, c + 1, row, column);
    dfs(grid, r - 1, c, row, column);
    dfs(grid, r, c - 1, row, column);
}

int main()
{
    int row, column, i, j, ans = 0;
    printf("row column:");
    scanf("%d %d", &row, &column);
    int** grid = (int**)malloc(sizeof(int*) * row);
    for (i = 0;i <= row - 1;i ++)
    {
        grid[i] = (int*)malloc(sizeof(int) * column);
    }
    
    for (i = 0;i <= row - 1;i ++)
    {
        for (j = 0;j <= column - 1;j ++)
        {
            scanf("%d", &grid[i][j]);
        }
    }

    for (i = 0; i <= row - 1;i ++)
    {
        for (j = 0;j <= column - 1;j ++)
        {
            if (grid[i][j] == 1)
            {
                dfs(grid, i, j, row, column);
                ans ++;
            }
        }
    }

    printf("\n%d", ans);
    return 0;
}