LeetCode-695：Max Area of Island （最大岛的面积）

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#LeetCode #算法 #DFS #最大岛面积

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本文介绍了一种使用深度优先搜索（DFS）算法解决二维数组中寻找最大岛屿面积的问题。通过递归的方式标记并计算每个岛屿的面积，最终找出最大值。

Question

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.

Note: The length of each dimension in the given grid does not exceed 50.

问题解析：

给定非空矩阵，元素为0或1，1代表陆地，四个方向中若有陆地相连，则代表区域，求最大区域的面积。

Answer

Solution 1：

DFS：Depth-first search，深度优先搜索。

用DFS算法实现每个区域的面积，这里利用递归实现，计算过的陆地则变为0；
遍历图中所有的元素，记录当前最大的面积。

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        int maxArea = 0;
        for (int i = 0; i < grid.length; i++){
            for (int j = 0; j < grid[0].length; j++){
                if (grid[i][j] != 0){
                    maxArea = Math.max(maxArea, areaOfIsland(grid, i, j));
                }
            }
        }

        return maxArea;
    }

    public int areaOfIsland(int[][] grid, int i, int j){
        if (i >= 0 && i < grid.length && j >= 0 && j < grid[0].length && grid[i][j] != 0){
            grid[i][j] = 0;
            return 1 + areaOfIsland(grid,i+1,j) + areaOfIsland(grid,i-1,j) + areaOfIsland(grid,i,j+1) + areaOfIsland(grid,i,j-1);
        }
        return 0;
    }
}