【easy】695. Max Area of Island

weixin_30639719發表於2020-04-05

 題目:

Given a non-empty 2D array grid of 0's and 1's, 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.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

[[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]]
Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.
//參考了答案……思路有,但是dfs寫的不對………………
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])  
                {  
                    res = max(res, dfs(grid, i, j));  
                }  
            }  
        }  
        return res;  
    }  
  
    int dfs(vector<vector<int>>& grid,int i,int j)  
    {  
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size()) return 0;  //超出邊界返回0
        if (grid[i][j])  
        {  
            grid[i][j] = 0;  
            return  1 + dfs(grid, i, j + 1) + dfs(grid, i + 1, j)  
                + dfs(grid, i, j - 1) + dfs(grid, i - 1, j);  //邊界內,且為1,返回XXX
        }  
        return 0;   //不是1 返回0
    } 
};

 

轉載於:https://www.cnblogs.com/sherry-yang/p/8283309.html

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