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leetcode | 695.岛屿的最大面积

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695.岛屿的最大面积

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

题目链接

思路

  1. 深度优先搜索（DFS）
    利用深度优先搜索（DFS）对每个岛屿进行遍历，找出最大值
    （1）从每个陆地出发，遍历该陆地所在的岛屿
    （2）在遍历某一岛屿的时候：
    1）从隶属于该岛屿的某一块陆地出发，向四个方向递归地DFS
    2）每次递归对下标进行判断，以区域的边界作为递归边界
    3）为保证每块陆地只访问一次，将已访问过的陆地置0
    4）递归地返回整块岛屿的面积
    （3）找出所有岛屿的最大值，即为答案

邻接矩阵和邻接表表示的图的DFS和BFS

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package test;

import java.util.LinkedList;
import java.util.Scanner;

class MGraph{
	private static final int MAXVEX = 100;
	private static final int INFINITY = 65535;
	
	int[] vexs = new int[MAXVEX];
	int[][] arc = new int[MAXVEX][MAXVEX];
	int numVertexes;
	int numEdges;
	boolean[] visited = new boolean[MAXVEX];
	
	public void createMGraph() {
		System.out.println("输入顶点数和边数：\n");
		Scanner scanner = new Scanner(System.in);
		numVertexes = scanner.nextInt();
		numEdges = scanner.nextInt();
		
		for(int i = 0; i < numVertexes; i++) {
			vexs[i] = scanner.nextInt();
		}
		
		for(int i = 0; i < numVertexes; i++) {
			for(int j = 0; j < numVertexes; j++) {
				arc[i][j] = INFINITY;
			}
		}
		
		for(int k = 0; k < numEdges; k++) {
			System.out.println("输入边（vi,vj）上的下标i,j和权w：\n");
			int i = scanner.nextInt();
			int j = scanner.nextInt();
			int w = scanner.nextInt();
			arc[i][j] = w;
			arc[j][i] = arc[i][j];
		}
	}
	
	public void DFS(int i) {
		 visited[i] = true;
		 System.out.println(vexs[i] + " ");
		 for(int j = 0; j < numVertexes; j++) {
			 if(arc[i][j] == 1 && !visited[j]) {
				 DFS(j);
			 }
		 }
	}
	
	public void DFSTraverse() {
		for(int i = 0; i < numVertexes; i++) {
			visited[i] = false;
		}
		
		for(int i = 0; i < numVertexes; i++) {
			if(!visited[i]) {
				DFS(i);
			}
		}
	}
	
	public void BFSTraverse() {
		LinkedList<Integer> queue = new LinkedList<Integer>();
		
		for(int i = 0; i < this.numVertexes; i++) {
			this.visited[i] = false;
		}
		
		for(int i = 0; i < this.numVertexes; i++) {//对每一个顶点做循环
			if(!visited[i]) {//若是未访问过就处理
				visited[i] = true;//设置当前顶点访问过
				System.out.println(this.vexs[i]);//打印顶点，也可以其他操作
				queue.addLast(i);//将此顶点入队列
				while(!queue.isEmpty()) {
					queue.pop();
					for(int j = 0; j < this.numVertexes; j++) {
						if(this.arc[i][j] == 1 && !this.visited[j]) {
							visited[j] = true;
							System.out.println(this.vexs[j]);
							queue.addLast(j);
						}
					}
				}
			}
		}
	}
}

class EdgeNode{
	int adjvex;
	EdgeNode next;
}

class VertexNode{
	int data;
	EdgeNode firstedge;
}

class GraphAdjList{
	private static final int MAXVEX = 100;
	VertexNode[] adjList= new VertexNode[MAXVEX];
	int numVertexes;
	int numEdges;
	boolean[] visited = new boolean[MAXVEX];
	
	public void createALGraph() {
		Scanner scanner = new Scanner(System.in);
		System.out.println("输入顶点数和边数：");
		this.numVertexes = scanner.nextInt();
		this.numEdges = scanner.nextInt();
		for(int i = 0 ; i < this.numVertexes; i++) {
			VertexNode node = new VertexNode();
			node.data = scanner.nextInt();
			node.firstedge = null;
			this.adjList[i] = node;
		}
		for(int k = 0; k < this.numEdges; k++) {
			System.out.println("输入边（vi,vj）上的顶点序号：\n");
			int i = scanner.nextInt();
			int j = scanner.nextInt();
			EdgeNode e = new EdgeNode();
			e.adjvex = j;
			e.next = this.adjList[i].firstedge;
			this.adjList[i].firstedge = e;
			
			EdgeNode e2 = new EdgeNode();
			e2.adjvex = i;
			e2.next = this.adjList[j].firstedge;
			this.adjList[j].firstedge = e2;
		}
	}
	
	public void DFS(int i) {
		visited[i] = true;
		System.out.println(this.adjList[i].data + " ");
		EdgeNode p = new EdgeNode();
		p = this.adjList[i].firstedge;
		while(p != null) {
			if(!visited[p.adjvex]) {
				DFS(p.adjvex);
			}
			p = p.next;
		}
	}
	
	public void DFSTraverse() {
		for(int i = 0; i < this.numVertexes; i++) {
			this.visited[i] = false;
		}
		
		for(int i = 0; i < this.numVertexes; i++) {
			if(!visited[i]) {
				this.DFS(i);
			}
		}
	}
	
	public void BFSTraverse() {
		LinkedList<Integer> queue = new LinkedList<Integer>();
		EdgeNode p = new EdgeNode();
		for(int i = 0; i < this.numVertexes; i++) {
			this.visited[i] = false;
		}
		for(int i = 0; i < this.numVertexes; i++) {
			if(!this.visited[i]) {
				this.visited[i] = true;
				System.out.println(this.adjList[i].data);//打印顶点
				queue.addLast(i);
				while(!queue.isEmpty()) {
					queue.pop();
					p = this.adjList[i].firstedge;
					while(p != null) {
						if(!this.visited[p.adjvex]) {
							this.visited[p.adjvex] = true;
							System.out.println(this.adjList[p.adjvex].data);
							queue.add(p.adjvex);
						}
						p = p.next;
					}
				}
			}
			
		}
	}
}

public class DFSAndBFS {
	public static void main(String[] args) {
		//邻接矩阵
		MGraph graph = new MGraph();
		graph.createMGraph();
		System.out.println("邻接矩阵：\n");
		for(int i = 0; i < graph.numVertexes; i++) {
			for(int j = 0; j < graph.numVertexes; j++) {
				System.out.print(graph.arc[i][j]+"\t");
			}
			System.out.println();
		}
		System.out.println("深度优先搜索遍历图（邻接矩阵）：\n");
		graph.DFSTraverse();
		System.out.println("广度优先搜索遍历图（邻接矩阵）：\n");
		graph.BFSTraverse();
		
		//邻接表
		GraphAdjList GL = new GraphAdjList();
		GL.createALGraph();
		System.out.println("深度优先搜索遍历图（邻接表）：\n");
		GL.DFSTraverse();
		System.out.println("广度优先搜索遍历图（邻接表）：\n");
		GL.BFSTraverse();
	}
}

代码

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class Solution:
	def island_dfs(self, grid, i, j):
		if len(grid) > i >= 0 and len(grid[0]) > j >= 0:
			if grid[i][j] == 0:
				return 0
			else:
				grid[i][j] = 0
				return 1 + self.island_dfs(grid, i - 1, j) + self.island_dfs(grid, i + 1, j) + \
				       self.island_dfs(grid, i, j - 1) + self.island_dfs(grid, i, j + 1)
		else:
			return 0
	
	def max_area_of_island(self, grid):
		ans = 0
		
		for i in range(len(grid)):
			for j in range(len(grid[0])):
				ans = max(ans, self.island_dfs(grid, i, j))
		
		return ans


if __name__ == "__main__":
	s = Solution()
	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]]
	max_area = s.max_area_of_island(grid)
	print(max_area)