Max Area of Island

最新推荐文章于 2022-05-13 15:04:41 发布

转载最新推荐文章于 2022-05-13 15:04:41 发布 · 205 阅读

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本文介绍了一个使用深度优先搜索(DFS)算法解决二维数组中寻找最大岛屿面积的问题。通过遍历每个元素并标记已访问区域来避免重复计算，最终找出最大岛屿的面积。

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Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

解析:就是给个二维数组,让我们找到最大的一块连着的区域,用深度优先搜索,判断一个区域的四周,看还有没有1,有的话就继续遍历四周,直到遍历完,

注意,每遍历一个就把它设为0,防止又遍历一遍.

import javax.swing.plaf.basic.BasicInternalFrameTitlePane.MaximizeAction;

public class Solution {
    public int maxAreaOfIsland(int[][] grid) {
    	int max_num = 0;
    	int result;
    	for(int i = 0; i < grid.length;i++)
    		for(int j = 0; j < grid[i].length; j++){
    			if(grid[i][j] == 1){
    				result = dfs(grid ,i,j);
    				
    				if(max_num < result){
        				max_num = result;
        			}		
    			}
    				
    		} 
    	
    	
       return max_num;
    }
    public int dfs(int[][] grid, int i, int j)
         {
             // if i or j is invalid or grid is 0, just return 0
             if( i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0)
                return 0;
             
             // do dfs to its 4 direction cell when value is 1
             int tempMaxArea = 1;    
             grid[i][j] = 0; // set current cell to 0 to prevent dfs coming back
            
           // order is left, top, right, bottom
             tempMaxArea += dfs(grid, i, j-1) + dfs(grid, i-1, j) + dfs(grid, i, j+1) + dfs(grid, i+1, j);
           
            return tempMaxArea;
    }
    public static void main(String args[]){
    	int[][] a = {{0,0,1,0,0,0,0,1,0,0,0,0,0},
    	             {0,0,0,0,0,0,0,1,1,1,0,0,0},
    	             {0,1,1,0,1,0,0,0,0,0,0,0,0},
    	             {0,1,0,0,1,1,0,0,1,0,1,0,0},
    	             {0,1,0,0,1,1,0,0,1,1,1,0,0},
    	             {0,0,0,0,0,0,0,0,0,0,1,0,0},
    	             {0,0,0,0,0,0,0,1,1,1,0,0,0},
    	             {0,0,0,0,0,0,0,1,1,0,0,0,0}};
    	Solution s = new Solution();
    	System.out.println(s.maxAreaOfIsland(a));
    	
    }
    
}