这道题让我学会了怎么把简单的题复杂化,一个简单题硬是让我复习了一遍DFS;其实参考评论,每遇到相邻的陆地,就让总周长减2即可,因为这两个陆地有重复的边,这条边不算入周长;
DFS:
int delta_x[4] = {0, 0, -1, 1};
int delta_y[4] = {-1, 1, 0, 0};
void dfs(vector<vector<int>> &grid, int &bordx, int &bordy, int x, int y, int &sum)
{
int c = 4;
for (int i = 0; i < 4; i++) //检查该土地的四个方位
{
int xx = x + delta_x[i], yy = y + delta_y[i];
if (xx < 0 || xx >= bordx || yy < 0 || yy >= bordy) //如果超出边界,直接下一个方位
{
continue;
}
if (grid[xx][yy]) //如果x,y周围xx,yy有陆地
{
c--; //该陆地周长减一
if (grid[xx][yy] == 1) //如果xx,yy陆地没有被计算过
{
grid[xx][yy] = 2; //置陆地xx,yy为计算过
dfs(grid, bordx, bordy, xx, yy, sum); //检差xx,yy陆地
}
}
}
sum += c;
}
int islandPerimeter(vector<vector<int>> &grid)
{
int len_row = grid.size();
if (len_row == 0)
{
return 0;
}
int len_col = grid[0].size();
int res = 0;
for (int i = 0, j = 0; i < len_row && j < len_col; j++)
{
if (grid[i][j] == 1) //找到岛屿入口
{
grid[i][j] = 2; //把入口置为计算过
dfs(grid, len_row, len_col, i, j, res);
break;
}
if (j == len_col - 1)
{
i++;
j = -1;
continue;
}
}
return res;
}
数学方法:
int islandPerimeter(vector<vector<int>> &grid)
{
//重点关注前面遍历过的方格,如果之前有相邻方格,就-2;
int rsp = 0;
int len_row = grid.size();
if (len_row == 0)
{
return 0;
}
int len_col = grid[0].size();
for (int i = 0; i < len_row; i++)
{
for (int j = 0; j < len_col; j++)
{
if (grid[i][j] == 1)
{
rsp += 4;
if (i > 0 && grid[i - 1][j] == 1)
{
rsp -= 2;
}
if (j > 0 && grid[i][j - 1] == 1)
{
rsp -= 2;
}
}
}
}
return rsp;
}