101.孤岛的总面积
基础题目 可以自己尝试做一做 。
代码随想录
方法1：深度搜索
import java.util.*;public class Main{public static int[][] dirc = {{0,1},{1,0},{0,-1},{-1,0}};public static void main (String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int m = scanner.nextInt();int[][] graph = new int[n][m];for(int i = 0; i < n ; i++){for(int j = 0; j < m; j++){graph[i][j] = scanner.nextInt();}}boolean[][] visited = new boolean[n][m];for(int i = 0; i < n; i++){if(graph[i][0] == 1){dfs(graph, visited, i , 0);}if(graph[i][m-1] == 1){dfs(graph, visited, i , m-1);}}for(int j = 0; j < m; j++){if(graph[0][j] == 1){dfs(graph, visited, 0, j);}if(graph[n-1][j] == 1){dfs(graph, visited, n-1, j);}}int sum = 0;for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(graph[i][j] == 1){sum ++;}}}System.out.println(sum);}public static void dfs(int[][] graph, boolean[][] visited, int x, int y){if(!(x >= 0 && x < graph.length && y >= 0 && y < graph[0].length )){return ;}if(visited[x][y] || graph[x][y] == 0){return ;}visited[x][y] = true;graph[x][y] = 0;for(int i = 0; i < dirc.length; i++){int nextX = x + dirc[i][0];int nextY = y + dirc[i][1];dfs(graph, visited, nextX, nextY);}}}方法2：广度搜索
import java.util.*;public class Main{public static int[][] dirc = {{0,1},{1,0},{0,-1},{-1,0}};public static void main (String[] args) {Scanner scanner = new Scanner(System.in);int n = scanner.nextInt();int m = scanner.nextInt();int[][] graph = new int[n][m];for(int i = 0; i < n ; i++){for(int j = 0; j < m; j++){graph[i][j] = scanner.nextInt();}}boolean[][] visited = new boolean[n][m];for(int i = 0; i < n; i++){if(graph[i][0] == 1){bfs(graph, visited, i , 0);}if(graph[i][m-1] == 1){bfs(graph, visited, i , m-1);}}for(int j = 0; j < m; j++){if(graph[0][j] == 1){bfs(graph, visited, 0, j);}if(graph[n-1][j] == 1){bfs(graph, visited, n-1, j);}}int sum = 0;for(int i = 0; i < n; i+&