【DFS】200.岛屿数量

Allenlzcoder

已于 2025-06-02 20:56:38 修改

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分类专栏：力扣Top100 文章标签：深度优先算法 dfs

于 2023-12-15 11:34:15 首次发布

本文链接：https://blog.youkuaiyun.com/Allenlzcoder/article/details/135013468

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该文章介绍了如何利用Java编程中的深度优先搜索(DFS)算法，通过遍历二维字符网格来计算其中的岛屿数量。Solution类中的`numIslands`方法为核心实现，`dfs`方法用于递归地探索每个岛屿。

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题目

Python

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        res = 0
        m, n = len(grid), len(grid[0])
        visited = [[0] * n for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1' and visited[i][j] == 0:
                    self.travel(i, j, grid, visited, m, n)
                    res += 1
        
        return res

    def travel(self, i, j, grid, visited, m, n):
        if (i < 0 or i >= m or j < 0 or j >= n
            or grid[i][j] == '0' or visited[i][j] == 1):
            return
        visited[i][j] = 1
        self.travel(i-1, j, grid, visited, m, n)
        self.travel(i+1, j, grid, visited, m, n)
        self.travel(i, j-1, grid, visited, m, n)
        self.travel(i, j+1, grid, visited, m, n)

Java

法1：岛屿数量

class Solution {
    public int numIslands(char[][] grid) {
        int m = grid.length, n = grid[0].length;
        int[][] used = new int[m][n];
        int res = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == '0' || used[i][j] == 1) {
                    continue;
                }
                ++res;
                dfs(i, j, grid, used, m, n);
            }
        }

        return res;
    }

    public void dfs(int i, int j, char[][] grid, int[][] used, int m, int n) {
        if (i < 0 || i >= m 
                || j < 0 || j >= n 
                || grid[i][j] == '0' || used[i][j] == 1) {
            return;
        }
        used[i][j] = 1;

        dfs(i - 1, j, grid, used, m, n); 
        dfs(i + 1, j, grid, used, m, n);
        dfs(i, j - 1, grid, used, m, n); 
        dfs(i, j + 1, grid, used, m, n);
    }
}