695. Max Area of Island. --- M

 You are given an 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.length
• n == grid[i].length
• 1 <= m, n <= 50
• grid[i][j] is either 0 or 1.

A:

就是经典的dfs

class Solution {

public:

int maxAreaOfIsland(vector<vector<int>>& grid) {

int res = 0;

for(int i =0; i<grid.size(); i++){

for(int j = 0 ;j < grid[0].size(); j++){

if(grid[i][j] == 1){

int curSize = 0;

dfs(grid, i, j, curSize);

res = max(res, curSize);

}

}

}

return res;

}

private:

void dfs(vector<vector<int>>& grid, int i, int j , int &curSize){

int m = grid.size(), n = grid[0].size();

if(i<0 || i >= m || j < 0 || j >= n || grid[i][j] == 0)

return;

grid[i][j] = 0;

curSize++;

vector<vector<int>> V = {{-1,0}, {1,0}, {0,1}, {0, -1}};

for(auto p : V){

dfs(grid, p[0] + i, p[1] + j, curSize );

}

}

};