694. Number of Distinct Islands 形状不同的岛屿数量

［抄题］：

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:

Given the above grid map, return 1.

Example 2:

Given the above grid map, return 3.

Notice that:

11
1

and

 1
11

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

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

不知道怎么存一块岛屿的形状：其实就是存方向构成的string就行了

每次走过一个点，都要设置该点为0，因为形状只统计一次，不重复

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［一句话思路］：

［输入量］：空：正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

DFS就是：走到底-到最后一行改变条件-没出界就再走到底，反正就是走到底就对了

　　　　　　　　　 {1, 1, 0},
                {0, 1, 1},
                {0, 0, 0},
                {1, 1, 1},
                {0, 1, 0}
为了分辨方向序列，加星号的位置不同。

curShape = rdr
curShape = rdr*
curShape = rdr**
curShape = rdr***
curShape = rd
curShape = rd*r
curShape = rd*r*
curShape = rd*r**

［一刷］：

stringbuilder需要专门的append函数而不是加号来连接string，所以在dfs中的参数就是它本身

［二刷］：

if grid[i][j] = 1后从主函数进去就行，dfs不用再写一次了

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

［复杂度］：Time complexity: O(mn) Space complexity: O(mn)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

[是否头一次写此类driver funcion的代码] ：

[潜台词] ：

class Solution {
    public int numDistinctIslands(int[][] grid) {
        //corner cases
        if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
        
        //initialization: StringBuilder, hashset
        StringBuilder curShape;
        Set<String> set = new HashSet<String>();
        
        //start to expand if grid[i][j] == 1
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    curShape = new StringBuilder();
                    
                    expandIslands(grid, i, j, curShape, "");
                    //add to set after trimed
                    set.add(curShape.toString().trim());
                }
            }
        }
        
        //return keyset
        return set.size();
    }
    
    public void expandIslands(int[][] grid, int x, int y, StringBuilder curShape, String directions) {
        //exit case
        if (grid[x][y] == 0 || x < 0 || x >= grid.length || y < 0 || y >= grid[0].length) return ;
        
            //set grid[x][y] = 0
            grid[x][y] = 0;
            //append the cur directions
            curShape.append(directins);
        
            //expand in 4 directions
            expandIslands(grid, x - 1, y, curShape, "U");
            expandIslands(grid, x + 1, y, curShape, "D");
            expandIslands(grid, x, y + 1, curShape, "R");
            expandIslands(grid, x, y - 1, curShape, "L");
            
            //add a * if neccessary
            curShape.append(" ");

        
    }
}

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