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0695-Max-Area-of-Island/Article/0695-Max-Area-of-Island.md
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# LeetCode 图解 |
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> 本文首发于公众号「图解面试算法」,是 [图解 LeetCode](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一。
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>
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> 同步博客:https://www.algomooc.com
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本题解作者:nettee
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## 题目描述
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给定一个包含了一些 `0` 和 `1` 的非空二维数组 `grid`。
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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。
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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`。)
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**示例 1:**
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```
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[[0,0,1,0,0,0,0,1,0,0,0,0,0],
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[0,0,0,0,0,0,0,1,1,1,0,0,0],
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[0,1,1,0,1,0,0,0,0,0,0,0,0],
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[0,1,0,0,1,1,0,0,1,0,1,0,0],
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[0,1,0,0,1,1,0,0,1,1,1,0,0],
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[0,0,0,0,0,0,0,0,0,0,1,0,0],
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[0,0,0,0,0,0,0,1,1,1,0,0,0],
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[0,0,0,0,0,0,0,1,1,0,0,0,0]]
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```
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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` 。
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**示例 2:**
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```
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[[0,0,0,0,0,0,0,0]]
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```
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对于上面这个给定的矩阵, 返回 0。
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注意: 给定的矩阵 `grid` 的长度和宽度都不超过 50。
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## 题目解析
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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。
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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。
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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。
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## 动画理解
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![](../Animation/Animation.gif)
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## 参考代码
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C++ 代码:
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```C++
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class Solution {
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public:
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int maxAreaOfIsland(vector<vector<int>>& grid) {
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int res = 0;
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for (int r = 0; r < grid.size(); r++) {
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for (int c = 0; c < grid[0].size(); c++) {
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if (grid[r][c] == 1) {
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int a = area(grid, r, c);
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res = max(res, a);
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}
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}
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}
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return res;
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}
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int area(vector<vector<int>>& grid, int r, int c) {
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if (!(inArea(grid, r, c))) {
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return 0;
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}
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if (grid[r][c] != 1) {
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return 0;
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}
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grid[r][c] = 2;
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return 1
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+ area(grid, r - 1, c)
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+ area(grid, r + 1, c)
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+ area(grid, r, c - 1)
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+ area(grid, r, c + 1);
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}
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bool inArea(vector<vector<int>>& grid, int r, int c) {
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return 0 <= r && r < grid.size()
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&& 0 <= c && c < grid[0].size();
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}
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};
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```
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Java 代码:
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```Java
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class Solution {
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public int maxAreaOfIsland(int[][] grid) {
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int res = 0;
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for (int r = 0; r < grid.length; r++) {
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for (int c = 0; c < grid[0].length; c++) {
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if (grid[r][c] == 1) {
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int a = area(grid, r, c);
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res = Math.max(res, a);
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}
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}
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}
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return res;
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}
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int area(int[][] grid, int r, int c) {
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if (!inArea(grid, r, c)) {
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return 0;
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}
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if (grid[r][c] != 1) {
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return 0;
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}
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grid[r][c] = 2;
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return 1
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+ area(grid, r - 1, c)
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+ area(grid, r + 1, c)
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+ area(grid, r, c - 1)
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+ area(grid, r, c + 1);
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}
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boolean inArea(int[][] grid, int r, int c) {
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return 0 <= r && r < grid.length
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&& 0 <= c && c < grid[0].length;
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}
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}
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```
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Python 代码:
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```Python
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class Solution:
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def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
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res = 0
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for r in range(len(grid)):
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for c in range(len(grid[0])):
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if grid[r][c] == 1:
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a = self.area(grid, r, c)
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res = max(res, a)
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return res
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def area(self, grid: List[List[int]], r: int, c: int) -> int:
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if not self.inArea(grid, r, c):
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return 0
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if grid[r][c] != 1:
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return 0
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grid[r][c] = 2
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return 1 \
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+ self.area(grid, r - 1, c) \
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+ self.area(grid, r + 1, c) \
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+ self.area(grid, r, c - 1) \
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+ self.area(grid, r, c + 1)
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def inArea(self, grid: List[List[int]], r: int, c: int) -> bool:
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return 0 <= r < len(grid) and 0 <= c < len(grid[0])
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```
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## 复杂度分析
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设网格的边长为 n,则时间复杂度为 O(n²)。
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