695. 岛屿的最大面积
给你一个大小为 m x n
的二进制矩阵 grid
。
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 水平或者竖直的四个方向上 相邻。你可以假设 grid
的四个边缘都被 0
(代表水)包围着。
岛屿的面积是岛上值为 1
的单元格的数目。
计算并返回 grid
中最大的岛屿面积。如果没有岛屿,则返回面积为 0
。
示例 1:
输入:grid = [[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]]
输出:6
解释:答案不应该是 11 ,因为岛屿只能包含水平或垂直这四个方向上的 1 。
示例 2:
输入:grid = [[0,0,0,0,0,0,0,0]]
输出:0
提示:
m == grid.length
n == grid[i].length
1 <= m, n <= 50
grid[i][j]
为0
或1
DFS
class Solution {
private:
const int dx[4] = {
0, 0, 1, -1};
const int dy[4] = {
-1, 1, 0, 0};
int dfs(vector<vector<int>>& grid, int x, int y) {
if (x < 0 || x >= grid.size() || y < 0 || y >= grid[0].size() || grid[x][y] == 0) return 0;
grid[x][y] = 0;
int area = 1;
area += dfs(grid, x, y + 1);
area += dfs(grid, x, y - 1);
area += dfs(grid, x - 1, y);
area += dfs(grid, x + 1, y);
return area;
}
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int maxArea = 0;
int n = grid.size(), m = grid[0].size();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == 1) {
maxArea = max(maxArea, dfs(grid, i, j));
}
}
}
return maxArea;
}
};
BFS
class Solution {
private:
const int dx[4] = {
0, 0, 1, -1};
const int dy[4] = {
-1, 1, 0, 0};
int bfs(vector<vector<int>>& grid, int x, int y) {
int area = 1;
queue<pair<int, int>> pointQueue;
pointQueue.emplace(x, y);
grid[x][y] = 0;
while (!pointQueue.empty()) {
int size = pointQueue.size();
for (int i = 0; i < size; ++i) {
auto point = pointQueue.front();
pointQueue.pop();
x = point.first;
y = point.second;
for (int j = 0; j < 4; ++j) {
int nx = x + dx[j];
int ny = y + dy[j];
if (nx >= 0 && nx < grid.size() && ny >= 0 && ny < grid[0].size() && grid[nx][ny] == 1) {
area += bfs(grid, nx, ny);
}
}
}
}
return area;
}
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int maxArea = 0;
int n = grid.size();
int m = grid[0].size();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (grid[i][j] == 1) {
maxArea = max(maxArea, bfs(grid, i, j));
}
}
}
return maxArea;
}
};