# LeetCode.695.Max Area of Island 岛屿的最大面积

## 题目描述

695 Max Area of Island
https://leetcode-cn.com/problems/max-area-of-island/

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.

## 解题过程

### DFS深度优先搜索-递归

private static class SolutionV2020 {
public int maxAreaOfIsland(int[][] grid) {
if (null == grid || 0 == grid.length) {
return 0;
}
int maxAres = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid.length; j++) {
if (grid[i][j] == 1) {
int area = dfs(grid, i, j);
maxAres = Math.max(maxAres, area);
}
}
}
return maxAres;
}

// 从i,j开始深度优先搜索，返回访问的1的个数
private int dfs(int[][] grid, int i, int j) {
int rows = grid.length, columns = grid.length;
int count = 0;
if (grid[i][j] == 1) {
grid[i][j] = 0;
count++;
if (i - 1 >= 0 && grid[i - 1][j] == 1) { // 上
count += dfs(grid, i - 1, j);
}
if (j + 1 < columns && grid[i][j + 1] == 1) { // 右
count += dfs(grid, i, j + 1);
}
if (i + 1 < rows && grid[i + 1][j] == 1) { // 下
count += dfs(grid, i + 1, j);
}
if (j - 1 >= 0 && grid[i][j - 1] == 1) { // 左
count += dfs(grid, i, j - 1);
}
}
return count;
}
}

## GitHub代码

algorithms/leetcode/leetcode/_695_MaxAreaOfIsland.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_695_MaxAreaOfIsland.java

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