題目描述：
m x n binary matrix grid. An island is a group of 1‘s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.
The area of an island is the number of cells with a value 1 in the island.
Return the maximum area of an island in grid. If there is no island, return 0.
Example 1:
Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
Output: 6
Explanation: The answer is not 11, because the island must be connected 4-directionally.
Example 2:
Input: grid = [[0,0,0,0,0,0,0,0]]
Output: 0
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 50grid[i][j]is either0or1.
題解（DFS）：
這題給了我們一個交錯數組，在這個數組中，1代表陸地，0代表海洋。如果有一片陸地互相接攘（陸地數>=1），並被海洋所包圍，則可稱為一個小島。問這個數組中，面積最大的島嶼面積大小，也就是數組中有最多相互連接的1的區域中，1的數量有多少。
這個題的解題思路是非常典型的深度搜索。創建一個isVisited的bool類型二維數組標志已經搜索過的區域，然後用兩重循環來遍歷整個交錯數組，對每個元素嘗試進行深度搜索。如果當前行列索引超出邊界/不是陸地（grid[row][col] == 0）/已經被搜索過，都直接返回。而找到陸地時，則對陸地的周邊區域通過遞歸進行DFS，直到檢查完所有相鄰的陸地，返回該「小島」的陸地面積。
然後在兩層循環中持續利用Math.Max來維護一個「最大陸地面積」的值，最後循環結束後返回的值就是本題的答案。
public class Solution {
public int MaxAreaOfIsland(int[][] grid) {
int row = grid.Length;
int col = grid[0].Length;
bool[,] isVisited = new bool[row, col];
int result = 0;
for(int y = 0; y < row; y++)
{
for(int x = 0; x < col; x++)
{
result = Math.Max(result, IslandCounter(grid, isVisited, y, x));
}
}
return result;
}
int IslandCounter(int[][] grid, bool[,] isVisited, int currRow, int currCol)
{
int rowEdge = grid.Length;
int colEdge = grid[0].Length;
//Edge Judge
if(currRow < 0 || currRow >= rowEdge || currCol < 0 || currCol >= colEdge) { return 0; }
if(grid[currRow][currCol] == 0) { return 0; }
//isVisited Judge
if(isVisited[currRow, currCol]) { return 0; }
//Make this be visited
isVisited[currRow, currCol] = true;
//DFS 4 dirs grid
//Up
int upCnt = IslandCounter(grid, isVisited, currRow - 1, currCol);
//Down
int downCnt = IslandCounter(grid, isVisited, currRow + 1, currCol);
//Left
int leftCnt = IslandCounter(grid, isVisited, currRow, currCol - 1);
//Right
int rightCnt = IslandCounter(grid, isVisited, currRow, currCol + 1);
return 1 + upCnt + downCnt + leftCnt + rightCnt;
}
}