Leetcode Island Perimeter

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本文介绍了一种使用深度优先搜索(DFS)算法来计算由方块构成的岛屿图形周长的方法。通过对每个陆地单元格进行DFS遍历，考虑边界条件，并通过加边去重的方式减少重复计算。

题意：找方块构成的图形的周长。

思路：DFS，注意边界条件。

class Solution {
public:
    vector<vector<bool> > ved;
    vector<vector<int> > mygrid;
    int xx, yy;
    int islandPerimeter(vector<vector<int>>& grid) {
        mygrid = grid;
        xx = grid.size();
        yy = grid[0].size();
        
        vector<bool> t(grid[0].size(), false);
        for(int i = 0; i < grid.size(); ++ i) ved.push_back(t);
        for(int i = 0; i < grid.size(); ++ i) {
            for(int j = 0; j < grid[i].size(); j ++) {
                if(grid[i][j] == 1) return dfs(i, j);
            }
        }
        return 0;
    }
    int dfs(int x, int y) {
        ved[x][y] = true;
        int p = 0;
        if(x - 1 >= 0) {
            if(ved[x - 1][y] == 0 && mygrid[x - 1][y] == 1) p += dfs(x - 1, y);
            if(mygrid[x - 1][y] == 0) p += 1;
        }
        if(x == 0) p += 1;
        
        if(y - 1 >= 0) {
            if(ved[x][y - 1] == 0 && mygrid[x][y - 1] == 1) p += dfs(x, y - 1);
            if(mygrid[x][y - 1] == 0) p += 1;
        }
        if(y == 0) p += 1;
        
        if(x + 1 < xx) {
            if(ved[x + 1][y] == 0 && mygrid[x + 1][y] == 1) p += dfs(x + 1, y);
            if(mygrid[x + 1][y] == 0) p += 1;
        }
        if(x == xx -1) p += 1;
        
        if(y + 1 < yy) {
            if(ved[x][y + 1] == 0 && mygrid[x][y + 1] == 1) p += dfs(x, y + 1);
            if(mygrid[x][y + 1] == 0) p += 1;
        }
        if(y == yy - 1) p += 1;
        
        //printf("x: %d, y: %d, p: %d\n", x, y, p);
        return p;
    }
};

另一种思路是加边去重。

    find how many 1 in the map. If without the consideration of surrounding cells, the total perimeter should be the total amount of 1 times 4.
    find how many cell walls that connect with both lands. We need to deduct twice of those lines from total perimeter