LeetCode 200. Number of Islands

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原创最新推荐文章于 2026-07-18 15:19:10 发布 1.9w 阅读 AI 写同款· GEO 优化 ›

leetcode解题之200. Number of Islands Java版（岛屿的数量） leetcode解题之200. Number of Islands Java版（岛屿的数量），深度优先遍历阅读详情

Given a 2d grid map of ‘1’s (land) and ‘0’s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.
Example 1:

Answer: 1
Example 2:

Answer: 3

题意：给你一个二维网图，‘1’代表陆地‘0’代表水。一个被水环绕的岛由水平或垂直邻接的陆地组成。假设网图的四个边被水环绕。
样例可以一眼就看出有多少个岛。

解题思路：我们可以访问每一个land，然后访问其周围的land，访问过后将land标记为water防止再次访问。

class Solution {
public:
    /*
        访问所有的点：
        1.遇到'0'(water)我们就返回
        2.遇到'1'(land),我们就把这个点设置成'0'(water),然后继续搜索其上下左右的点

        Custom testcase:
        [["1","1","1"],["0","1","0"],["1","1","1"]] 让我Wrong Answer了一次(只访问了下边和右边)
    */
    int numIslands(vector<vector<char>>& grid) {
        if(grid.empty()) return 0;
        int rows = grid.size(), cols = grid[0].size();

        int lands = 0;  // land总数
        // 遍历每一个点，遇到新的land时，++lands
        for(int r = 0; r < rows; ++r){
            for(int c = 0; c < cols; ++c){
                if(grid[r][c] == '1'){
                    searchLand(r, c, grid);
                    ++lands;
                }
            }
        }
        return lands;
    }

private:
    void searchLand(int row, int col, vector<vector<char>>& grid){
        if(row >= grid.size() || col >= grid[0].size() || row < 0 || col < 0) return;   // 边界
        if(grid[row][col] == '0') return;   // 遇到了water

        grid[row][col] = '0';  // 将该节点设置为'0'
        searchLand(row - 1, col, grid);    // 访问上边的点
        searchLand(row + 1, col, grid);    // 访问下边的点
        searchLand(row, col - 1, grid);    // 访问左边的点
        searchLand(row, col + 1, grid);    // 访问右边的点
    }
};