[Leetcode] 711. Number of Distinct Islands II 解题报告

LeetCode 711：独特岛屿的数量解题策略

最新推荐文章于 2024-01-15 10:35:03 发布

原创最新推荐文章于 2024-01-15 10:35:03 发布 · 4.7k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#Leetcode #解题报告 #DFS

IT公司面试习题专栏收录该内容

736 篇文章

订阅专栏

这篇博客探讨了LeetCode上的711题，即计算非空二维数组中不同岛屿的数量。岛屿是由1组成的四向连接的群体，四周被0包围。岛屿被认为是相同的，如果它们形状相同，或者经过90、180或270度旋转或水平/垂直翻转。博主通过DFS方法解决此问题，重点在于处理岛屿的旋转和反射。每发现一个岛屿，会生成其8种变形，并选择排序最小的形式作为代表。解决方案借鉴了694题的思路。

题目：

Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Count the number of distinct islands. An island is considered to be the same as another if they have the same shape, or have the same shape after rotation (90, 180, or 270 degrees only) or reflection (left/right direction or up/down direction).

Example 1:

Given the above grid map, return 1 .

Notice that:

11
1

and

 1
11

are considered same island shapes. Because if we make a 180 degrees clockwise rotation on the first island, then two islands will have the same shapes.

Example 2:

Given the above grid map, return 2 .

Here are the two distinct islands:

111
1

and

1
1

Notice that:

111
1

and

1
111

are considered same island shapes. Because if we flip the first array in the up/down direction, then they have the same shapes.

Note: The length of each dimension in the given grid does not exceed 50.

思路：

DFS的套路不难，这里主要是如何处理小岛的旋转问题。我们的做法是：每次找到一个小岛之后，我们生成其8个对应的变形（包括本身，翻转以及旋转等）。然后返回其排序最小的那个，作为这种类型的小岛的代表。这样其余的思路就都可以照搬 [Leetcode] 694. Number of Distinct Islands 解题报告中的思路了。

代码：

class Solution {
public:
    int numDistinctIslands2(vector<vector<int>>& grid) {
        int row_num = grid.size(), col_num = grid[0].size();
        set<vector<pair<int, int>>> islands;
        for (int r = 0; r < row_num; ++r) {
            for (int c = 0; c < col_num; ++c) {
                if (grid[r][c] == 1) {
                    vector<pair<int, int>> island;
                    DFS(grid, r, c, r, c, island);
                    islands.insert(normalize(island));
                }
            }
        }
        return islands.size();
    }
private:
    void DFS(vector<vector<int>> &grid, int r0, int c0, int r, int c, vector<pair<int, int>> &island) {
        int row_num = grid.size(), col_num = grid[0].size();
        if (r < 0 || r >= row_num || c < 0 || c >= col_num || grid[r][c] <= 0) {
            return;
        }
        grid[r][c] = -1;
        island.push_back(make_pair(r - r0, c - c0));
        for (int d = 0; d < 4; ++d) {
            DFS(grid, r0, c0, r + delta[d][0], c + delta[d][1], island);
        }
    }
    vector<pair<int, int>> normalize(vector<pair<int, int>> &a) {
        vector<vector<pair<int, int>>> ret(8, vector<pair<int, int>>());
        for (auto &p : a) {
            int x = p.first, y = p.second;
            ret[0].push_back(make_pair(x, y));
            ret[1].push_back(make_pair(x, -y));
            ret[2].push_back(make_pair(-x, y));
            ret[3].push_back(make_pair(-x, -y));
            ret[4].push_back(make_pair(y, x));
            ret[5].push_back(make_pair(y, -x));
            ret[6].push_back(make_pair(-y, x));
            ret[7].push_back(make_pair(-y, -x));
        }
        for (int i = 0; i < 8; ++i) {
            sort(ret[i].begin(), ret[i].end());
            int r_offset = 0 - ret[i][0].first, c_offset = 0 - ret[i][0].second;
            for (int j = 0; j < ret[i].size(); ++j) {
                ret[i][j].first += r_offset;
                ret[i][j].second += c_offset;
            }
        }
        sort(ret.begin(), ret.end());
        return ret[0];
    } 
    int delta[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
};