孤岛问题算法题

  1 package com.guava;
  2 
  3 import java.util.Scanner;
  4 
  5 public class Main2 {
  6 
  7     private int m = 0, n = 0;
  8 
  9     int res = 0;
 10 
 11     public static void main(String[] args) {
 12         Scanner sc = new Scanner(System.in);
 13         String[] in = sc.nextLine().split(" ");
 14         int m = Integer.parseInt(in[0]);
 15         int n = Integer.parseInt(in[1]);
 16         int i = 0;
 17         int[][] grid = new int[m][n];
 18         while (i < m) {
 19             if (sc.hasNext()) {
 20                 String[] arr = sc.nextLine().split(" ");
 21                 int j = 0;
 22                 while (j < n) {
 23                     grid[i][j] = Integer.parseInt(arr[j]);
 24                     j++;
 25                 }
 26             }
 27             i++;
 28         }
 29 
 30         Main2 main = new Main2();
 31         System.out.println(main.getNumIslands(grid));
 32 
 33     }
 34 
 35     //方法一：会改变输入的二维数组
 36     public int getNumIslands(int[][] grid) {
 37         m = grid.length;
 38         n = grid[0].length;
 39         if (m == 0) {
 40             return 0;
 41         }
 42 
 43         if (n == 0) {
 44             return 0;
 45         }
 46 
 47         for (int i = 0; i < m; i++) {
 48             for (int j = 0; j < n; j++) {
 49                 if (grid[i][j] != 1) continue;
 50                 res++;
 51                 //回溯法
 52                 calculateIsland(grid, i, j);
 53             }
 54         }
 55         return res;
 56     }
 57 
 58     public void calculateIsland(int[][] grid, int i, int j) {
 59         if (i < 0 || i >= m || j < 0 || j >= n) {
 60             return;
 61         }
 62         if (grid[i][j] == 1) {
 63             grid[i][j] = 0;
 64             calculateIsland(grid, i - 1, j);
 65             calculateIsland(grid, i + 1, j);
 66             calculateIsland(grid, i, j - 1);
 67             calculateIsland(grid, i, j + 1);
 68         }
 69     }
 70 
 71 
 72     //方法二：使用与输入的二位数组相同大小的二维数组visit，来保存是否被遍历到。
 73     public int getLands(char[][] lands) {
 74         if (lands.length == 0) {
 75             return 0;
 76         }
 77         int count = 0;
 78         int row = lands.length;
 79         int col = lands[0].length;
 80         int[][] visit = new int[row][col];
 81         for (int i = 0; i < row; i++) {
 82             for (int j = 0; j < col; j++) {
 83                 if (visit[i][j] != 1 && lands[i][j] == ‘1‘) //注意此处的判断条件
 84                     count++;
 85                 //回溯法
 86                 BFS(lands, i, j, visit);
 87             }
 88         }
 89 
 90         return count;
 91     }
 92 
 93     private void BFS(char[][] lands, int i, int j, int[][] visit) {
 94         if (i < 0 || i >= lands.length || j < 0 || j >= lands[0].length) {
 95             return;
 96         }
 97         if (lands[i][j] == ‘0‘) {
 98             return;
 99         }
100 
101         if (lands[i][j] == ‘1‘) {
102             visit[i][j] = ‘0‘;
103             BFS(lands, i - 1, j, visit);
104             BFS(lands, i + 1, j, visit);
105             BFS(lands, i, j - 1, visit);
106             BFS(lands, i, j + 1, visit);
107         }
108     }
109 }