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；
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；
