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695. 岛屿的最大面积

English Version

题目描述

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 。

 

示例 1：

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 。

示例 2：

输入：grid = [[0,0,0,0,0,0,0,0]]
输出：0

 

提示：

  • m == grid.length
  • n == grid[i].length
  • 1 <= m, n <= 50
  • grid[i][j] 为 0 或 1

解法

方法一：DFS

方法二：并查集

并查集模板：

模板 1——朴素并查集：

# 初始化，p存储每个点的父节点
p = list(range(n))

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]


# 合并a和b所在的两个集合
p[find(a)] = find(b)

模板 2——维护 size 的并查集：

# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
p = list(range(n))
size = [1] * n

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        # 路径压缩
        p[x] = find(p[x])
    return p[x]

# 合并a和b所在的两个集合
if find(a) != find(b):
    size[find(b)] += size[find(a)]
    p[find(a)] = find(b)

模板 3——维护到祖宗节点距离的并查集：

# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
p = list(range(n))
d = [0] * n

# 返回x的祖宗节点
def find(x):
    if p[x] != x:
        t = find(p[x])
        d[x] += d[p[x]]
        p[x] = t
    return p[x]

# 合并a和b所在的两个集合
p[find(a)] = find(b)
d[find(a)] = distance

Python3

DFS：

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        def dfs(i, j):
            grid[i][j] = 0
            ans = 1
            for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
                x, y = i + a, j + b
                if 0 <= x < m and 0 <= y < n and grid[x][y] == 1:
                    ans += dfs(x, y)
            return ans

        m, n = len(grid), len(grid[0])
        return max([dfs(i, j) for i in range(m) for j in range(n) if grid[i][j] == 1], default=0)

并查集：

class Solution:
    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        m, n = len(grid), len(grid[0])
        p = list(range(m * n))
        size = [1] * (m * n)
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    for a, b in [[0, 1], [1, 0]]:
                        x, y = i + a, j + b
                        if 0 <= x < m and 0 <= y < n and grid[x][y] == 1 and find(i * n + j) != find(x * n + y):
                            size[find(x * n + y)] += size[find(i * n + j)]
                            p[find(i * n + j)] = find(x * n + y)
        return max([size[i * n + j] for i in range(m) for j in range(n) if grid[i][j] == 1], default=0)

Java

DFS：

class Solution {
    private int[][] grid;
    private int m;
    private int n;

    public int maxAreaOfIsland(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ans = Math.max(ans, dfs(i, j));
                }
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        grid[i][j] = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        int ans = 1;
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k];
            int y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                ans += dfs(x, y);
            }
        }
        return ans;
    }
}

并查集：

class Solution {
    private int[] p;
    private int[] size;

    public int maxAreaOfIsland(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        p = new int[m * n];
        size = new int[m * n];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
            size[i] = 1;
        }
        int[] dirs = {0, 1, 0};
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    for (int k = 0; k < 2; ++k) {
                        int x = i + dirs[k];
                        int y = j + dirs[k + 1];
                        if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1
                            && find(i * n + j) != find(x * n + y)) {
                            size[find(x * n + y)] += size[find(i * n + j)];
                            p[find(i * n + j)] = find(x * n + y);
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ans = Math.max(ans, size[i * n + j]);
                }
            }
        }
        return ans;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

TypeScript

DFS：

function maxAreaOfIsland(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let ans = 0;
    const dirs = [-1, 0, 1, 0, -1];
    let dfs = function (i, j) {
        grid[i][j] = 0;
        let ans = 1;
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1) {
                ans += dfs(x, y);
            }
        }
        return ans;
    };
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == 1) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
    }
    return ans;
}

并查集：

function maxAreaOfIsland(grid: number[][]): number {
    const m = grid.length;
    const n = grid[0].length;
    let p = new Array(m * n);
    for (let i = 0; i < p.length; ++i) {
        p[i] = i;
    }
    let size = new Array(m * n).fill(1);
    let find = function (x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };
    const dirs = [1, 0, 1];
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == 1) {
                for (let k = 0; k < 2; ++k) {
                    const x = i + dirs[k];
                    const y = j + dirs[k + 1];
                    if (
                        x >= 0 &&
                        x < m &&
                        y >= 0 &&
                        y < n &&
                        grid[x][y] == 1 &&
                        find(x * n + y) != find(i * n + j)
                    ) {
                        size[find(x * n + y)] += size[find(i * n + j)];
                        p[find(i * n + j)] = find(x * n + y);
                    }
                }
            }
        }
    }
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] == 1) {
                ans = Math.max(ans, size[i * n + j]);
            }
        }
    }
    return ans;
}

C++

DFS：

class Solution {
public:
    int m;
    int n;

    int maxAreaOfIsland(vector<vector<int>>& grid) {
        m = grid.size();
        n = grid[0].size();
        int ans = 0;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (grid[i][j] == 1)
                    ans = max(ans, dfs(i, j, grid));
        return ans;
    }

    int dfs(int i, int j, vector<vector<int>>& grid) {
        grid[i][j] = 0;
        int ans = 1;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k];
            int y = j + dirs[k + 1];
            if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1)
                ans += dfs(x, y, grid);
        }
        return ans;
    }
};

并查集：

class Solution {
public:
    vector<int> p;
    vector<int> size;

    int maxAreaOfIsland(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        p.resize(m * n);
        size.resize(m * n, 1);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        vector<int> dirs = {0, 1, 0};
        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                if (grid[i][j])
                {
                    for (int k = 0; k < 2; ++k)
                    {
                        int x = i + dirs[k];
                        int y = j + dirs[k + 1];
                        if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] && find(i * n + j) != find(x * n + y))
                        {
                            size[find(x * n + y)] += size[find(i * n + j)];
                            p[find(i * n + j)] = find(x * n + y);
                        }
                    }
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (grid[i][j])
                    ans = max(ans, size[i * n + j]);
        return ans;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

Go

DFS：

func maxAreaOfIsland(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	dirs := []int{-1, 0, 1, 0, -1}
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		grid[i][j] = 0
		ans := 1
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 {
				ans += dfs(x, y)
			}
		}
		return ans
	}
	ans := 0
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				ans = max(ans, dfs(i, j))
			}
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

并查集：

func maxAreaOfIsland(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	p := make([]int, m*n)
	size := make([]int, m*n)
	for i := range p {
		p[i] = i
		size[i] = 1
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	dirs := []int{1, 0, 1}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				for k := 0; k < 2; k++ {
					x, y := i+dirs[k], j+dirs[k+1]
					if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 && find(i*n+j) != find(x*n+y) {
						size[find(x*n+y)] += size[find(i*n+j)]
						p[find(i*n+j)] = find(x*n + y)
					}
				}
			}
		}
	}
	ans := 0
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				ans = max(ans, size[i*n+j])
			}
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

Rust

DFS：

impl Solution {
    fn dfs(grid: &mut Vec<Vec<i32>>, i: usize, j: usize) -> i32 {
        if i == grid.len() || j == grid[0].len() || grid[i][j] == 0 {
            return 0;
        }
        grid[i][j] = 0;
        let mut res = 1 + Self::dfs(grid, i + 1, j) + Self::dfs(grid, i, j + 1);
        if i != 0 {
            res += Self::dfs(grid, i - 1, j)
        }
        if j != 0 {
            res += Self::dfs(grid, i, j - 1)
        }
        res
    }

    pub fn max_area_of_island(mut grid: Vec<Vec<i32>>) -> i32 {
        let m = grid.len();
        let n = grid[0].len();
        let mut res = 0;
        for i in 0..m {
            for j in 0..n {
                res = res.max(Self::dfs(&mut grid, i, j))
            }
        }
        res
    }
}

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