【LeetCode】695. Max Area of Island(岛的最大面积)

本文针对LeetCode上的题目MaxAreaofIsland进行了详细的解析,通过深度优先搜索(DFS)的方法来寻找二维数组中由1表示的岛屿的最大面积。

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【LeetCode】695. Max Area of Island(岛的最大面积)

题目链接:https://leetcode.com/problems/max-area-of-island/description/

难度:Easy

题目描述:

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.)

Example 1:

[[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]]
Given the above grid, return  6 . Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

解释:找到上下左右中,1最多的那块的数量。

思路:DFS(深度优先搜索)

深度优先遍历图的方法是,从图中某顶点v出发:
(1)访问顶点v;
(2)依次从v的未被访问的邻接点出发,对图进行深度优先遍历;直至图中和v有路径相通的顶点都被访问;
(3)若此时图中尚有顶点未被访问,则从一个未被访问的顶点出发,重新进行深度优先遍历,直到图中所有顶点均被访问过为止。

代码实现:C++同Java

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        int max=0;
        for(int i=0;i<grid.length;i++)
        {
            for(int j=0;j<grid[0].length;j++)
            {
                if(1==grid[i][j])
                    max=Integer.max(max,DfsApply(grid,i,j));
            }
        }
        return max;
    }
    public int DfsApply(int[][]grid,int i,int j)
    {
        if(i>=0&&i<grid.length&&j>=0&&j<grid[0].length&&grid[i][j]==1)
        {
            grid[i][j]=0;
            return 1+DfsApply(grid,i+1,j)+DfsApply(grid,i-1,j)+DfsApply(grid,i,j+1)+DfsApply(grid,i,j-1);
        }
        else return 0;
    }
}



日期:2018/2/23-23:04

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