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# LeetCode 图解 |
> 本文首发于公众号图解面试算法 [图解 LeetCode](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一
>
> 同步博客https://www.algomooc.com
本题解作者nettee
## 题目描述
给定一个包含了一些 `0` `1` 的非空二维数组 `grid`
一个**岛屿**是由一些相邻的 `1` (代表土地) 构成的组合这里的相邻要求两个 `1` 必须在水平或者竖直方向上相邻你可以假设 `grid` 的四个边缘都被 `0`代表水包围着
找到给定的二维数组中最大的岛屿面积(如果没有岛屿则返回面积为 `0`)
**示例 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]]
```
对于上面这个给定矩阵应返回 6注意答案不应该是 11 因为岛屿只能包含水平或垂直的四个方向的 `1`
**示例 2:**
```
[[0,0,0,0,0,0,0,0]]
```
对于上面这个给定的矩阵, 返回 0
注意: 给定的矩阵 `grid` 的长度和宽度都不超过 50
## 题目解析
这道题的主要思路是深度优先搜索每次走到一个是 1 的格子就搜索整个岛屿并计算当前岛屿的面积最后返回岛屿面积的最大值
网格可以看成是一个无向图的结构每个格子和它上下左右的四个格子相邻如果四个相邻的格子坐标合法且是陆地就可以继续搜索
在深度优先搜索的时候要注意避免重复遍历我们可以把已经遍历过的陆地改成 2这样遇到 2 我们就知道已经遍历过这个格子了不进行重复遍历
## 动画理解
![](../Animation/Animation.gif)
## 参考代码
C++ 代码
```C++
class Solution {
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int res = 0;
for (int r = 0; r < grid.size(); r++) {
for (int c = 0; c < grid[0].size(); c++) {
if (grid[r][c] == 1) {
int a = area(grid, r, c);
res = max(res, a);
}
}
}
return res;
}
int area(vector<vector<int>>& grid, int r, int c) {
if (!(inArea(grid, r, c))) {
return 0;
}
if (grid[r][c] != 1) {
return 0;
}
grid[r][c] = 2;
return 1
+ area(grid, r - 1, c)
+ area(grid, r + 1, c)
+ area(grid, r, c - 1)
+ area(grid, r, c + 1);
}
bool inArea(vector<vector<int>>& grid, int r, int c) {
return 0 <= r && r < grid.size()
&& 0 <= c && c < grid[0].size();
}
};
```
Java 代码
```Java
class Solution {
public int maxAreaOfIsland(int[][] grid) {
int res = 0;
for (int r = 0; r < grid.length; r++) {
for (int c = 0; c < grid[0].length; c++) {
if (grid[r][c] == 1) {
int a = area(grid, r, c);
res = Math.max(res, a);
}
}
}
return res;
}
int area(int[][] grid, int r, int c) {
if (!inArea(grid, r, c)) {
return 0;
}
if (grid[r][c] != 1) {
return 0;
}
grid[r][c] = 2;
return 1
+ area(grid, r - 1, c)
+ area(grid, r + 1, c)
+ area(grid, r, c - 1)
+ area(grid, r, c + 1);
}
boolean inArea(int[][] grid, int r, int c) {
return 0 <= r && r < grid.length
&& 0 <= c && c < grid[0].length;
}
}
```
Python 代码
```Python
class Solution:
def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
res = 0
for r in range(len(grid)):
for c in range(len(grid[0])):
if grid[r][c] == 1:
a = self.area(grid, r, c)
res = max(res, a)
return res
def area(self, grid: List[List[int]], r: int, c: int) -> int:
if not self.inArea(grid, r, c):
return 0
if grid[r][c] != 1:
return 0
grid[r][c] = 2
return 1 \
+ self.area(grid, r - 1, c) \
+ self.area(grid, r + 1, c) \
+ self.area(grid, r, c - 1) \
+ self.area(grid, r, c + 1)
def inArea(self, grid: List[List[int]], r: int, c: int) -> bool:
return 0 <= r < len(grid) and 0 <= c < len(grid[0])
```
## 复杂度分析
设网格的边长为 n则时间复杂度为 O(n²)