695. Max Area of Island

问题描述

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.)

Example 1:

[[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.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.
Note: The length of each dimension in the given grid does not exceed 50.

题目链接：

---

思路分析

在一个二维数组表示的图中，1表示陆地，找的连起来的陆地的最大面积。

正好和自己的wumpus world 的移动方式相同，直接跑BFS就可以了。

代码

class Solution {
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int area = 0;
        for (int i = 0; i < grid.size(); i++){
            for (int j = 0; j < grid[i].size(); j++){
                if (grid[i][j] == 1)
                    area = max(area, BFS(i, j, grid));
            }
        }
        return area;
    }

    int BFS(int i, int j, vector<vector<int>>& grid){
        int count = 1;
        grid[i][j] = 2;
        if (i < grid.size() - 1 && grid[i+1][j] == 1){
            count += BFS(i+1, j, grid);
        }
        if (i > 0 && grid[i-1][j] == 1){
            count += BFS(i-1, j, grid);
        }
        if (j < grid[i].size() - 1 && grid[i][j+1] == 1){
            count += BFS(i, j+1, grid);
        }
        if (j > 0 && grid[i][j-1] == 1){
            count += BFS(i, j-1, grid);
        }
        return count;

    }
};

时间复杂度：$O(r*c)$
空间复杂度：$O(r*c)$

---

反思

一开始是有些懵逼的，不过很快的反应过来了。自己的二维数组表示迷宫的方式果然是有些落伍的啊。