题目:463. 岛屿的周长
思路
深度优先搜索;
代码
Method 1
对于遍历到的一个地块,向四周探索,越界或者遇到海洋地块说明这条边需要统计;
cpp
class Solution {
private:
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int result = 0; // 周长
void dfs(vector<vector<int>>& grid, vector<vector<bool>>& vistied, int x, int y)
{
int m = grid.size(), n = grid[0].size();
int i, j;
int next_x, next_y;
for(i = 0; i < 4; i++)
{
next_x = x + dir[i][0];
next_y = y + dir[i][1];
if(next_x < 0 || next_x >= m || next_y < 0 || next_y >= n)
{
result++;
continue;
}
if(grid[next_x][next_y] == 0)
{
result++;
}
if(!vistied[next_x][next_y] && grid[next_x][next_y] == 1)
{
vistied[next_x][next_y] = true;
dfs(grid, vistied, next_x, next_y);
}
}
}
public:
int islandPerimeter(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
int i, j;
vector<vector<bool>> vistied(m, vector<bool>(n, false));
for(i = 0; i < m; i++)
{
for(j = 0; j < n; j++)
{
if(!vistied[i][j] && grid[i][j] == 1)
{
vistied[i][j] = true;
dfs(grid, vistied, i, j);
}
}
}
return result;
}
};
Method 2
初始周长 = 岛屿地块 * 4,在岛屿内部,有一对相邻地块,周长-2
;