Description

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
11000
00011
00011
Given the above grid map, return 1.
Example 2:
11011
10000
00001
11011
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.
Note: The length of each dimension in the given grid does not exceed 50.

Problem URL

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Solution

給一二維數組表示1為島嶼，找到不同形狀的島嶼，旋轉相同也不可以的。是一個典型的dfs題目，雙for循環遍歷，如果是1（很關鍵）則開始dfs操作，在dfs中，如果越界或者不為1或者visited就返回，然后對四個方向進行dfs操作，最后設置對正常的返回值（回退情況）。模版題。

Using a string to denote the shape of an island. If the island have same shape, the dfs traversal order would be same order. Remember add ‘b’ for back track.

Code

class Solution {public int numDistinctIslands(int[][] grid) {if (grid == null || grid.length == 0 || grid[0].length == 0){return 0;}Set<String> set = new HashSet<>();for (int i = 0; i < grid.length; i++){for (int j = 0; j < grid[i].length; j++){if (grid[i][j] != 0){StringBuilder sb = new StringBuilder();dfs(grid, sb, i, j, 'o');set.add(sb.toString()); }}}return set.size();}private void dfs(int[][] grid, StringBuilder sb, int i, int j, char dir){if (i < 0 || i >= grid.length || j < 0 || j >= grid[i].length || grid[i][j] == 0){return;}sb.append(dir);grid[i][j] = 0;dfs(grid, sb, i + 1, j, 'd');dfs(grid, sb, i - 1, j, 'u');dfs(grid, sb, i, j + 1, 'r');dfs(grid, sb, i, j - 1, 'l');sb.append('b');} }

Time Complexity: O(mn)
Space Complexity: O(mn)

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Review

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