LeetCode #694 - Number of Distinct Islands

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本文介绍了一个算法问题，即在一个由0和1组成的二维数组中，如何计算不同岛屿的数量。岛屿定义为一组4方向相连的1（代表陆地），且不考虑旋转或反射。通过深度优先搜索（DFS）遍历每个岛屿，并将岛屿的相对坐标记录为pair数组，使用set进行去重，最终返回不同岛屿的数量。

题目描述：

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 and only if one island can be translated (and not rotated or reflected) to equal the other.

Example 1:

11000

00011

Given the above grid map, return 1.

Example 2:

11011

10000

00001

11011

Given the above grid map, return 3.

Notice that:

and

are considered different island shapes, because we do not consider reflection / rotation.

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

class Solution {
public:
    int numDistinctIslands(vector<vector<int>>& grid) {
        if(grid.size()==0||grid[0].size()==0) return 0;
        int m=grid.size();
        int n=grid[0].size();
        set<vector<pair<int,int>>> s; // 用set来去重，pair数组表示整个连通区域的每个节点相对于第一个节点的相对坐标，如果两个连通区域形状相同，而且由于DFS遍历顺序固定，得到的相对坐标数组必定相同。
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]==1) 
                {
                    vector<pair<int,int>> v;
                    DFS(grid,i,j,i,j,v);
                    s.insert(v);
                }
            }
        }
        return s.size();
    }
    
    void DFS(vector<vector<int>>& grid, int x0, int y0, int i, int j, vector<pair<int,int>>& v)
    {
        if(grid[i][j]==0) return;
        grid[i][j]=0;
        v.push_back({i-x0,j-y0});
        if(i>0) DFS(grid,x0,y0,i-1,j,v);
        if(i<grid.size()-1) DFS(grid,x0,y0,i+1,j,v);
        if(j>0) DFS(grid,x0,y0,i,j-1,v);
        if(j<grid[0].size()-1) DFS(grid,x0,y0,i,j+1,v);
    }
};