200. 岛屿数量

题目描述

给你一个由 1(陆地)和0(水)组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且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。

此外，你可以假设该网格的四条边均被水包围。

输入输出

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输入：grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
输出：1

输入：grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
输出：3

基本思路

深度优先搜索 经过的数组修改为2 时复$O(MN)$ 空复$O(MN)$

广度优先搜索 ：

  1. 为了求出岛屿的数量，我们可以扫描整个二维网格。如果一个位置为 1，则将其加入队列，开始进行广度优先搜索。在广度优先搜索的过程中，每个搜索到的 1 都会被重新标记为 1。直到队列为空，搜索结束。

  2. 最终岛屿的数量就是我们进行广度优先搜索的次数。

  3. 时复$O(MN)$ 空复$O(min(M, N))$ 空复:最坏情况下均为陆地 队列大小的值

java实现

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// 深搜 会修改二维数组中的值
class Solution {
    public int result;
    public int numIslands(char[][] grid) {
        result = 0;
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == '1'){
                    dfs(grid, i, j);
                    result++;
                }
            }
        }
        return result;
    }

    private void dfs(char[][] grid, int row, int col){
        if(row >=  grid.length || col >= grid[0].length || row < 0 || col < 0){
            return;
        }
        if(grid[row][col] != '1'){
            return;
        }
        grid[row][col] = '2';
        dfs(grid, row-1, col);
        dfs(grid, row+1, col);
        dfs(grid, row, col-1);
        dfs(grid, row, col+1);

    }
}

// 深搜 不会修改数组中的值 new一个visited[][]解决
class Solution {
    public int numIslands(char[][] grid) {
        int num_island = 0;
        int[][] visited = new int[grid.length][grid[0].length];
        for(int i = 0; i < grid.length; i++) {
            for(int j = 0; j < grid[0].length; j++) {
                if(grid[i][j] == '1' && visited[i][j] != 1){
                    dfs(grid, i, j, visited);
                    num_island++;
                }
            }
        }
        return num_island;
    }

    private void dfs(char[][] grid, int row, int col, int[][] visited){
        if(row >=  grid.length || col >= grid[0].length || row < 0 || col < 0
                || grid[row][col] == '0' || visited[row][col] == 1){
            return;
        }
        
        visited[row][col] = 1;
        dfs(grid, row-1, col, visited);
        dfs(grid, row+1, col, visited);
        dfs(grid, row, col-1, visited);
        dfs(grid, row, col+1, visited);

    }
}