![](https://img2020.cnblogs.com/blog/2146166/202010/2146166-20201030103143628-323770052.png)
- 每有一个位置是1,则周长+4
- 如果上一个是1,则公用上边,周长-2;如果左一个是1,则公用左边,周长-2
class Solution {
public:
int islandPerimeter(vector<vector<int>>& grid) {
int n = grid.size();
if(n==0) return 0;
int m = grid[0].size();
int sum = 0;
for(int i=0; i<n; ++i){
for(int j=0; j<m; ++j){
if(grid[i][j] == 1){
sum += 4;
//上
if(i>0 && grid[i-1][j]==1){
sum -= 2;
}
//左
if(j>0 && grid[i][j-1]==1){
sum -= 2;
}
}
}
}
return sum;
}
};
![](https://img2020.cnblogs.com/blog/2146166/202010/2146166-20201030111300008-2075629281.png)
- 先找起点,也就是
grid[i][j]==1
- 从起点开始dfs, 上下左右四个方向,每有一个方向上
grid[i][j]==1
,则岛屿面积+1
- 为防止重复遍历,需要记录访问过的地方。方法一:原地修改数组,访问过的地方置0; 方法二:额外数组标记访问过的状态
class Solution {
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
n = grid.size();
if(n==0) return 0;
m = grid[0].size();
vector<vector<bool>> visited(n, vector<bool>(m,0));
for(int i=0; i<n; ++i){
for(int j=0; j<m; ++j){
if(!visited[i][j] && grid[i][j]==1){
maxArea = max(maxArea, dfs(grid, visited, i, j));
}
}
}
return maxArea;
}
int dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int row, int col){
if(row<0 || row>=n || col<0 || col>=m || grid[row][col] == 0) return 0;
//
//cout << row << "," << col << "=" << grid[row][col] << "\n";
int cnt = 0;
if(!visited[row][col]){
cnt = 1;
visited[row][col] = true;
cnt += dfs(grid, visited, row-1, col);
cnt += dfs(grid, visited, row+1, col);
cnt += dfs(grid, visited, row, col-1);
cnt += dfs(grid, visited, row, col+1);
}
return cnt;
}
private:
int maxArea = 0;
int n, m;
};