岛屿数量-LeetCode200

原创已于 2022-04-15 14:47:05 修改 · 115 阅读

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#java #深度优先遍历

于 2022-04-15 14:46:24 首次发布

本文解析了如何使用深度优先搜索算法计数给定二维网格中被水包围的陆地区域，通过实例和代码展示了如何标记未访问过的陆地并计算岛屿数目。适合理解二维空间上的图论应用。

一、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。

示例一：

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

示例二：

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

二、解题思路

深度优先遍历：设置isVisited[]数组标记二维数组中当前位置的元素是否被访问；利用二重循环对二维数组进行遍历，如果当前的grid[i][j]='1'且isVisited[i][j]=false，则以当前元素为起点进行深度优先遍历（广度优先遍历也可），进行dfs的次数即为“岛屿”数。

三、代码实现

class Solution {
    public int numIslands(char[][] grid) {
        int row = grid.length;
        int col = grid[0].length;
        //标记是否被访问过
        boolean[][] isVisited = new boolean[row][col];
        //记录岛屿数
        int islandCnt = 0;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (grid[i][j] == '1' && !isVisited[i][j]) {
                    islandCnt++;
                    //以其为中心进行深度优先遍历
                    dfs(grid, isVisited, i, j);
                }
            }
        }
        return islandCnt;
    }

    public void dfs(char[][] grid, boolean[][] isVisited, int i, int j) {
        int row = grid.length;
        int col = grid[0].length;
        if ( i == row || j == col || i == -1 || j == -1|| grid[i][j] =='0'||isVisited[i][j]) {
            return;
        }
        isVisited[i][j]=true;
        //深度优先遍历时，右、下、左、上四个方向
        int[][] directions = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
        for (int m = 0; m < 4; m++) {
            int rowIndex = i + directions[m][0];
            int colIndex = j + directions[m][1];
            dfs(grid, isVisited, rowIndex, colIndex);
        }
    }
}