200. Number of Islands

yPhantom 2019年10月14日 40次浏览

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

Solution

找到2d数组中"岛"的数量。这题类似于之前写过的1219. Path with Maximum Gold。都是用的dfs,直接贴代码吧:

class Solution {
    
    private final int[] d = {0, 1, 0, -1, 0};
    
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length < 1 || grid[0].length < 1) {
            return 0;
        }
        int count = 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);
                    count++;
                }
            }
        }
        return count;
    }
    
    private void dfs(char[][] grid, int i, int j) {
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] != '1') {
            return;
        }
        grid[i][j] = '0';
        for(int k = 0; k < 4; k++) {
            dfs(grid, i + d[k], j + d[k + 1]);
        }
    }
}

基本上看到这种2d的字眼,就是bfs,dfs,dp三种算法了。