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# 695. 岛屿的最大面积

力扣题目链接 (opens new window)

给你一个大小为 m x n 的二进制矩阵 grid 。

岛屿是由一些相邻的 1 (代表土地) 构成的组合，这里的「相邻」要求两个 1 必须在水平或者竖直的四个方向上相邻。你可以假设 grid 的四个边缘都被 0（代表水）包围着。

岛屿的面积是岛上值为 1 的单元格的数目。

计算并返回 grid 中最大的岛屿面积。如果没有岛屿，则返回面积为 0 。

输入：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]]
输出：6
解释：答案不应该是 11 ，因为岛屿只能包含水平或垂直这四个方向上的 1 。

# 思路

注意题目中每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

也就是说斜角度链接是不算了，例如示例二，是三个岛屿，如图：

这道题目也是 dfs bfs基础类题目，就是搜索每个岛屿上“1”的数量，然后取一个最大的。

本题思路上比较简单，难点其实都是 dfs 和 bfs的理论基础，关于理论基础我在这里都有详细讲解：

# DFS

很多同学，写dfs其实也是凭感觉来，有的时候dfs函数中写终止条件才能过，有的时候 dfs函数不写终止添加也能过！

这里其实涉及到dfs的两种写法。

写法一，dfs只处理下一个节点，即在主函数遇到岛屿就计数为1，dfs处理接下来的相邻陆地

// 版本一
class Solution {
private:
    int count;
    int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
    void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
        for (int i = 0; i < 4; i++) {
            int nextx = x + dir[i][0];
            int nexty = y + dir[i][1];
            if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;  // 越界了，直接跳过
            if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { // 没有访问过的 同时 是陆地的

                visited[nextx][nexty] = true;
                count++;
                dfs(grid, visited, nextx, nexty);
            }
        }
    }

public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));

        int result = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!visited[i][j] && grid[i][j] == 1) {
                    count = 1;  // 因为dfs处理下一个节点，所以这里遇到陆地了就先计数，dfs处理接下来的相邻陆地
                    visited[i][j] = true;
                    dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
                    result = max(result, count);
                }
            }
        }
        return result;
    }
};

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写法二，dfs处理当前节点，即即在主函数遇到岛屿就计数为0，dfs处理接下来的全部陆地

dfs

// 版本二
class Solution {
private:
    int count;
    int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
    void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
        if (visited[x][y] || grid[x][y] == 0) return; // 终止条件：访问过的节点 或者 遇到海水
        visited[x][y] = true; // 标记访问过
        count++;
        for (int i = 0; i < 4; i++) {
            int nextx = x + dir[i][0];
            int nexty = y + dir[i][1];
            if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;  // 越界了，直接跳过
            dfs(grid, visited, nextx, nexty);
        }
    }

public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
        int result = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!visited[i][j] && grid[i][j] == 1) {
                    count = 0; // 因为dfs处理当前节点，所以遇到陆地计数为0，进dfs之后在开始从1计数
                    dfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
                    result = max(result, count);
                }
            }
        }
        return result;
    }
};

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大家通过注释可以发现，两种写法，版本一，在主函数遇到陆地就计数为1，接下来的相邻陆地都在dfs中计算。版本二在主函数遇到陆地计数为0，也就是不计数，陆地数量都去dfs里做计算。

这也是为什么大家看了很多，dfs的写法，发现写法怎么都不一样呢？其实这就是根本原因。

以上两种写法的区别，我在题解： DFS，BDF 你没注意的细节都给你列出来了！LeetCode：200. 岛屿数量 (opens new window)做了详细介绍。

# BFS

关于广度优先搜索，如果大家还不了解的话，看这里：广度优先搜索精讲 (opens new window)

本题BFS代码如下：

class Solution {
private:
    int count;
    int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
    void bfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
        queue<int> que;
        que.push(x);
        que.push(y);
        visited[x][y] = true; // 加入队列就意味节点是陆地可到达的点
        count++;
        while(!que.empty()) {
            int xx = que.front();que.pop();
            int yy = que.front();que.pop();
            for (int i = 0 ;i < 4; i++) {
                int nextx = xx + dir[i][0];
                int nexty = yy + dir[i][1];
                if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界
                if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) { // 节点没有被访问过且是陆地
                    visited[nextx][nexty] = true;
                    count++;
                    que.push(nextx);
                    que.push(nexty);
                }
            }
        }
    }

public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector<bool>> visited = vector<vector<bool>>(n, vector<bool>(m, false));
        int result = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!visited[i][j] && grid[i][j] == 1) {
                    count = 0;
                    bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
                    result = max(result, count);
                }
            }
        }
        return result;
    }
};

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# 其它语言版本

# Java

# DFS

// DFS
class Solution {
    int[][] dir = {
        {0, 1}, //right
        {1, 0}, //down
        {0, -1}, //left
        {-1, 0} //up
    };
    boolean visited[][];
    int count;
    public int maxAreaOfIsland(int[][] grid) {
        int res = 0;
        visited = new boolean[grid.length][grid[0].length];
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(visited[i][j] == false && grid[i][j] == 1){
                    count = 0;
                    dfs(grid, i, j);
                    res = Math.max(res, count);
                }
            }
        }
        return res;
    }
    private void dfs(int[][] grid, int x, int y){
        if(visited[x][y] == true || grid[x][y] == 0)
            return;
        
        visited[x][y] = true;
        count++;

        for(int i = 0; i < 4; i++){
            int nextX = x + dir[i][0];
            int nextY = y + dir[i][1];

            if(nextX < 0 || nextY < 0 || nextX >= grid.length || nextY >= grid[0].length)
                continue;
            dfs(grid, nextX, nextY);
        }
    }
}

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# BFS

//BFS
class Solution {
    int[][] dir = {
        {0, 1}, {1, 0}, {0, -1}, {-1, 0}
    };

    int count;
    boolean visited[][];

    public int maxAreaOfIsland(int[][] grid) {
        int res = 0;
        visited = new boolean[grid.length][grid[0].length];
        
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(visited[i][j] == false && grid[i][j] == 1){
                    count = 0;
                    bfs(grid, i, j);
                    res = Math.max(res, count);
                }
            }
        }
        return res;
    }
    private void bfs(int[][] grid, int x, int y){
        Queue<Integer> que = new LinkedList<>();
        que.offer(x);
        que.offer(y);
        visited[x][y] = true;
        count++;
        
        while(!que.isEmpty()){
            int currX = que.poll();
            int currY = que.poll();

            for(int i = 0; i < 4; i++){
                int nextX = currX + dir[i][0];
                int nextY = currY + dir[i][1];
                
                if(nextX < 0 || nextY < 0 || nextX >= grid.length || nextY >= grid[0].length)
                    continue;
                if(visited[nextX][nextY] == false && grid[nextX][nextY] == 1){
                    que.offer(nextX);
                    que.offer(nextY);
                    visited[nextX][nextY] = true;
                    count++;
                }
            }
        }
    }
}

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# DFS 優化（遇到島嶼後，就把他淹沒）

//这里使用深度优先搜索 DFS 来完成本道题目。我们使用 DFS 计算一个岛屿的面积，同时维护计算过的最大的岛屿面积。同时，为了避免对岛屿重复计算，我们在 DFS 的时候对岛屿进行 “淹没” 操作，即将岛屿所占的地方置为 0。
public int maxAreaOfIsland(int[][] grid) {
    int res = 0;
    for(int i = 0;i < grid.length;i++){
        for(int j = 0;j < grid[0].length;j++){
        	//每遇到一个岛屿就计算这个岛屿的面积同时”淹没“这个岛屿
            if(grid[i][j] == 1){
            	//每次计算一个岛屿的面积都要与res比较，维护最大的岛屿面积作为最后的答案
                res = Math.max(res,dfs(grid,i,j));
            }
        }
    }
    return res;
}
public int dfs(int[][] grid,int i,int j){
	//搜索边界：i，j超过grid的范围或者当前元素为0，即当前所在的地方已经是海洋
    if(i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0) return 0;
    //淹没土地，防止后续被重复计算
    grid[i][j] = 0;
    //递归的思路：要求当前土地(i,j)所在的岛屿的面积，则等于1加上下左右相邻的土地的总面积
    return 1 + dfs(grid,i - 1,j) +
               dfs(grid,i + 1,j) +
               dfs(grid,i,j + 1) +
               dfs(grid,i,j - 1);
}

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# Python

# BFS

class Solution:
    def __init__(self):
        self.count = 0

    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        # 与200.独立岛屿不同的是：此题grid列表内是int！！！

        # BFS
        if not grid: return 0

        m, n = len(grid), len(grid[0])
        visited = [[False for i in range(n)] for j in range(m)]

        result = 0
        for i in range(m):
            for j in range(n):
                if not visited[i][j] and grid[i][j] == 1:
                    # 每一个新岛屿
                    self.count = 0
                    self.bfs(grid, visited, i, j)
                    result = max(result, self.count)

        return result

    def bfs(self, grid, visited, i, j):
        self.count += 1
        visited[i][j] = True    

        queue = collections.deque([(i, j)])
        while queue:
            x, y = queue.popleft()
            for new_x, new_y in [(x + 1, y), (x - 1, y), (x, y - 1), (x, y + 1)]:
                if 0 <= new_x < len(grid) and 0 <= new_y < len(grid[0]) and not visited[new_x][new_y] and grid[new_x][new_y] == 1:
                    visited[new_x][new_y] = True
                    self.count += 1
                    queue.append((new_x, new_y))

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# DFS

class Solution:
    def __init__(self):
        self.count = 0

    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        # DFS
        if not grid: return 0

        m, n = len(grid), len(grid[0])
        visited = [[False for _ in range(n)] for _ in range(m)]

        result = 0
        for i in range(m):
            for j in range(n):
                if not visited[i][j] and grid[i][j] == 1:
                    self.count = 0
                    self.dfs(grid, visited, i, j)
                    result = max(result, self.count)
        return result
        
    def dfs(self, grid, visited, x, y):
        if visited[x][y] or grid[x][y] == 0:
            return 
        visited[x][y] = True
        self.count += 1
        for new_x, new_y in [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]:
            if 0 <= new_x < len(grid) and 0 <= new_y < len(grid[0]): 
                self.dfs(grid, visited, new_x, new_y)

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# JavaScript

var maxAreaOfIsland = function (grid) {
    let dir = [[0, 1], [1, 0], [-1, 0], [0, -1]]; // 四个方向
    
    let visited = new Array(grid.length).fill().map(() => Array(grid[0].length).fill(false))
    
    let dfs = (grid, visited, x, y, m) => {
        for (let i = 0; i < 4; i++) {
            let nextX = x + dir[i][0]
            let nextY = y + dir[i][1]
            if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length)
                continue;
            if (!visited[nextX][nextY] && grid[nextX][nextY] === 1) {
                visited[nextX][nextY] = true
                m = dfs(grid, visited, nextX, nextY,m+1)
            }
        }
        return m
    }

    let  max = 0

    for (let i = 0; i < grid.length; i++) {
        for (let j = 0; j < grid[i].length; j++) {
            if (!visited[i][j] && grid[i][j] === 1) {
                // 深度优先
                visited[i][j] = true;
                let m = dfs(grid, visited, i, j, 1);
                if (m > max) max = m;
            }
        }
    }
    return max
};

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# Go

dsf: 版本一

var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
var count int = 0

func maxAreaOfIsland(grid [][]int) int {
	res := 0
	visited := make([][]bool, len(grid))
	for i := 0; i < len(grid); i++ {
		visited[i] = make([]bool, len(grid[0]))
	}

	for i, rows := range grid {
		for j, v := range rows {
			if v == 1 && !visited[i][j] {
				// 第一种写法，重制 count，必定有 1 个
				count = 1
				dfs(grid, visited, i, j)
				res = max(res, count)
			}

		}
	}

	return res
}

// 第一种写法
func dfs(grid [][]int, visited [][]bool, i, j int) {
	visited[i][j] = true // 标记已访问，循环中未标记会导致重复
	for _, d := range DIRECTIONS {
		x, y := i+d[0], j+d[1]
		if x < 0 || x >= len(grid) || y < 0 || y >= len(grid[0]) {
			continue
		}
		if grid[x][y] == 1 && !visited[x][y] {
			count++
			dfs(grid, visited, x, y)
		}
	}
}

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dfs：版本二

var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
var count int = 0

func maxAreaOfIsland(grid [][]int) int {
	res := 0
	visited := make([][]bool, len(grid))
	for i := 0; i < len(grid); i++ {
		visited[i] = make([]bool, len(grid[0]))
	}

	for i, rows := range grid {
		for j, v := range rows {
			if v == 1 && !visited[i][j] {
				// 第二种写法
				count = 0
				dfs(grid, visited, i, j)
				res = max(res, count)
			}

		}
	}

	return res
}

// 第二种写法
func dfs(grid [][]int, visited [][]bool, i, j int) {
	if visited[i][j] || grid[i][j] == 0 {
		return
	}
	visited[i][j] = true
	count++
	for _, d := range DIRECTIONS {
		x, y := i+d[0], j+d[1]
		if x < 0 || x >= len(grid) || y < 0 || y >= len(grid[0]) {
			continue
		}
		dfs(grid, visited, x, y)
	}
}

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bfs:

var DIRECTIONS = [4][2]int{{-1, 0}, {0, -1}, {1, 0}, {0, 1}}
var count int = 0

func maxAreaOfIsland(grid [][]int) int {
	res := 0
	visited := make([][]bool, len(grid))
	for i := 0; i < len(grid); i++ {
		visited[i] = make([]bool, len(grid[0]))
	}

	for i, rows := range grid {
		for j, v := range rows {
			if v == 1 && !visited[i][j] {
				count = 0
				bfs(grid, visited, i, j)
				res = max(res, count)
			}

		}
	}

	return res
}

// bfs
func bfs(grid [][]int, visited [][]bool, i, j int) {
	visited[i][j] = true
	count++
	queue := [][2]int{{i, j}}
	for len(queue) > 0 {
		cur := queue[0]
		queue = queue[1:]
		for _, d := range DIRECTIONS {
			x, y := cur[0]+d[0], cur[1]+d[1]
			if x < 0 || x >= len(grid) || y < 0 || y >= len(grid[0]) {
				continue
			}
			if grid[x][y] == 1 && !visited[x][y] {
				count++
				queue = append(queue, [2]int{x, y})
				visited[x][y] = true
			}
		}
	}
}

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# Rust

dfs: 版本一

impl Solution {
    const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)];

    pub fn max_area_of_island(grid: Vec<Vec<i32>>) -> i32 {
        let mut visited = vec![vec![false; grid[0].len()]; grid.len()];

        let mut res = 0;
        for (i, nums) in grid.iter().enumerate() {
            for (j, &num) in nums.iter().enumerate() {
                if !visited[i][j] && num == 1 {
                    let mut count = 1;
                    visited[i][j] = true;
                    Self::dfs(&grid, &mut visited, (i as i32, j as i32), &mut count);
                    res = res.max(count);
                }
            }
        }

        res
    }

    pub fn dfs(
        grid: &[Vec<i32>],
        visited: &mut [Vec<bool>],
        (x, y): (i32, i32),
        count: &mut i32,
    ) {
        for (dx, dy) in Self::DIRECTIONS {
            let (nx, ny) = (x + dx, y + dy);
            if nx < 0 || nx >= grid.len() as i32 || ny < 0 || ny >= grid[0].len() as i32 {
                continue;
            }
            let (nx, ny) = (nx as usize, ny as usize);
            if !visited[nx][ny] && grid[nx][ny] == 1 {
                visited[nx][ny] = true;
                *count += 1;
                Self::dfs(grid, visited, (nx as i32, ny as i32), count);
            }
        }
    }
}

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dfs: 版本二

impl Solution {
    const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)];

    pub fn max_area_of_island(grid: Vec<Vec<i32>>) -> i32 {
        let mut visited = vec![vec![false; grid[0].len()]; grid.len()];

        let mut res = 0;
        for (i, nums) in grid.iter().enumerate() {
            for (j, &num) in nums.iter().enumerate() {
                if !visited[i][j] && num == 1 {
                    let mut count = 0;
                    Self::dfs(&grid, &mut visited, (i as i32, j as i32), &mut count);
                    res = res.max(count);
                }
            }
        }

        res
    }

    pub fn dfs(
        grid: &[Vec<i32>],
        visited: &mut [Vec<bool>],
        (x, y): (i32, i32),
        count: &mut i32,
    ) {
        if visited[x as usize][y as usize] || grid[x as usize][y as usize] == 0 {
            return;
        }
        visited[x as usize][y as usize] = true;
        *count += 1;
        for (dx, dy) in Self::DIRECTIONS {
            let (nx, ny) = (x + dx, y + dy);
            if nx < 0 || nx >= grid.len() as i32 || ny < 0 || ny >= grid[0].len() as i32 {
                continue;
            }
            Self::dfs(grid, visited, (nx, ny), count);
        }
    }
}

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bfs:

use std::collections::VecDeque;
impl Solution {
    const DIRECTIONS: [(i32, i32); 4] = [(0, 1), (1, 0), (-1, 0), (0, -1)];

    pub fn max_area_of_island(grid: Vec<Vec<i32>>) -> i32 {
        let mut visited = vec![vec![false; grid[0].len()]; grid.len()];

        let mut res = 0;
        for (i, nums) in grid.iter().enumerate() {
            for (j, &num) in nums.iter().enumerate() {
                if !visited[i][j] && num == 1 {
                    let mut count = 0;
                    Self::bfs(&grid, &mut visited, (i as i32, j as i32), &mut count);
                    res = res.max(count);
                }
            }
        }

        res
    }

    pub fn bfs(grid: &[Vec<i32>], visited: &mut [Vec<bool>], (x, y): (i32, i32), count: &mut i32) {
        let mut queue = VecDeque::new();
        queue.push_back((x, y));
        visited[x as usize][y as usize] = true;
        *count += 1;
        while let Some((cur_x, cur_y)) = queue.pop_front() {
            for (dx, dy) in Self::DIRECTIONS {
                let (nx, ny) = (cur_x + dx, cur_y + dy);
                if nx < 0 || nx >= grid.len() as i32 || ny < 0 || ny >= grid[0].len() as i32 {
                    continue;
                }
                let (nx, ny) = (nx as usize, ny as usize);
                if !visited[nx][ny] && grid[nx][ny] == 1 {
                    visited[nx][ny] = true;
                    queue.push_back((nx as i32, ny as i32));
                    *count += 1;
                }
            }
        }
    }
}

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