LeetCode——695.岛屿的最大面积

给定一个包含了一些 0 和 1的非空二维数组 grid , 一个 岛屿 是由四个方向 (水平或垂直) 的 1 (代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。

找到给定的二维数组中最大的岛屿面积。(如果没有岛屿，则返回面积为0。)

示例 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]]

对于上面这个给定矩阵应返回 6。注意答案不应该是11，因为岛屿只能包含水平或垂直的四个方向的‘1’。

示例 2:

[[0,0,0,0,0,0,0,0]]
对于上面这个给定的矩阵, 返回 0。

注意: 给定的矩阵grid 的长度和宽度都不超过 50。

https://leetcode-cn.com/problems/max-area-of-island

递归

遍历grid，当遇到为1的点，我们调用递归函数，

在递归函数中，我们首先判断i和j是否越界，还有grid[i][j]是否为1，我们没有用visited数组，而是直接修改了grid数组，遍历过的标记为-1。

如果合法，那么cnt自增1，并且更新结果res，然后对其周围四个相邻位置分别调用递归函数即可，参见代码如下：

c++

class Solution {
public:
    vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size(), res = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] != 1) continue;
                int cnt = 0;
                helper(grid, i, j, cnt, res);
            }
        }
        return res;
    }
    void helper(vector<vector<int>>& grid, int i, int j, int& cnt, int& res) {
        int m = grid.size(), n = grid[0].size();
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] <= 0) return;
        res = max(res, ++cnt);
        grid[i][j] *= -1;
        for (auto dir : dirs) {
            helper(grid, i + dir[0], j + dir[1], cnt, res);
        }
    }
};

迭代

BFS遍历，使用queue来辅助运算，思路没啥太大区别，都是套路，都是模版，往里套就行了，参见代码如下：

c++

class Solution {
public:
    vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size(), res = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] != 1) continue;
                int cnt = 0;
                queue<pair<int, int>> q{{{i, j}}};
                grid[i][j] *= -1;
                while (!q.empty()) {
                    auto t = q.front(); q.pop();
                    res = max(res, ++cnt);
                    for (auto dir : dirs) {
                        int x = t.first + dir[0], y = t.second + dir[1];
                        if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] <= 0) continue;
                        grid[x][y] *= -1;
                        q.push({x, y});
                    }
                }
            }
        }
        return res;
    }
};

java

class Solution {

    public int maxAreaOfIsland(int[][] grid) {
        Deque<int[]> queue = new LinkedList<>();

        int[][] moveIndexArray = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        int maxArea = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[i].length; j++) {
                queue.add(new int[]{i, j});
                //计算最大面积
                int currMaxArea = 0;
                while (!queue.isEmpty()) {
                    int size = queue.size();
                    for (int k = 0; k < size; k++) {
                        int[] poll = queue.poll();
                        int currI = poll[0];
                        int currJ = poll[1];
                        if (currI < 0 || currI >= grid.length || currJ < 0 || currJ >= grid[0].length || grid[currI][currJ] == 0) {
                            continue;
                        }
                        currMaxArea++;
                        grid[currI][currJ] = 0;
                        for (int[] moveIndex : moveIndexArray) {
                            queue.offer(new int[]{currI + moveIndex[0], currJ + moveIndex[1]});
                        }
                    }
                }
                maxArea = Math.max(currMaxArea, maxArea);
            }
        }
        return maxArea;
    }
}