200NumberofIslands.org

[#B] 200 Number of Islands - LeetCode

题目

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:

Input:
11110
11010
11000
00000

Output: 1

Example 2:

Input:
11000
11000
00100
00011

Output: 3

思路：

通过深度遍历，把每个 ‘1’ 周围的 ‘1’ 都置为 ‘0’ 。

Java

class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0)
            return 0;
        int numOfIslands = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    numOfIslands++;
                    dfs(grid, i, j);
                }
            
            }
        }
        return numOfIslands;
    }
 
    public void dfs(char[][] grid, int r, int c) {
        if (r < 0 || r >= grid.length || c < 0 || c >= grid[0].length)
            return;
        if (grid[r][c] == '1'){
            grid[r][c] = '0';
            dfs(grid, r - 1, c);
            dfs(grid, r, c + 1);
            dfs(grid, r + 1, c);
            dfs(grid, r, c - 1);
        }
     
    }
}

CPP

class Solution {
public:
 
    int d[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
    int m,n;
    int numIslands(vector<vector<char>>& grid) {
        m = grid.size();
        if (m == 0)
            return 0;
        n = grid[0].size();
        if (n == 0)
            return 0;
        int islands = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == '1') {
                    islands++;
                    dfs(grid, i, j);
                }
            }
        }
        return islands;
    }
 
    void dfs(vector<vector<char>>& grid, int x, int y) {
        if (grid[x][y] == '1') {
            grid[x][y] = '0';
            for (int i = 0; i < 4; i++) {
                int newX = x + d[i][0];
                int newY = y + d[i][1];
                if (inArea(newX, newY) && grid[newX][newY] == '1') {
                    dfs(grid, newX, newY);
                }
            }
        }
    }
 
    bool inArea(int x, int y) {
        return x >= 0 && x < m && y >= 0 && y < n;
    }
};