Open
Description
In a N x N grid composed of 1 x 1 squares, each 1 x 1 square consists of a /, \, or blank space. These characters divide the square into contiguous regions.
(Note that backslash characters are escaped, so a \ is represented as "\\".)
Return the number of regions.
Example 1:
Input: [ " /", "/ " ] Output: 2 Explanation: The 2x2 grid is as follows: 
Example 2:
Input: [ " /", " " ] Output: 1 Explanation: The 2x2 grid is as follows: 
Example 3:
Input: [ "\\/", "/\\" ] Output: 4 Explanation: (Recall that because \ characters are escaped, "\\/" refers to \/, and "/\\" refers to /\.) The 2x2 grid is as follows: 
Example 4:
Input: [ "/\\", "\\/" ] Output: 5 Explanation: (Recall that because \ characters are escaped, "/\\" refers to /\, and "\\/" refers to \/.) The 2x2 grid is as follows: 
Example 5:
Input: [ "//", "/ " ] Output: 3 Explanation: The 2x2 grid is as follows: 
Note:
1 <= grid.length == grid[0].length <= 30grid[i][j]is either'/','\', or' '.
这道题说是有个 NxN 个小方块，每个小方块里可能是斜杠，反斜杠，或者是空格。然后问这些斜杠能将整个区域划分成多少个小区域。这的确是一道很有意思的题目，虽然只是 Medium 的难度，但是博主拿到题目的时候是懵逼的，这尼玛怎么做？无奈只好去论坛上看大神们的解法，结果发现大神们果然牛b，巧妙的将这道题转化为了岛屿个数问题 Number of Islands，具体的做法将每个小区间化为九个小格子，这样斜杠或者反斜杠就是对角线或者逆对角线了，是不是有点图像像素化的感觉，就是当你把某个图片尽可能的放大后，到最后你看到也就是一个个不同颜色的小格子组成了这幅图片。这样只要把斜杠的位置都标记为1，而空白的位置都标记为0，这样只要找出分隔开的0的群组的个数就可以了，就是岛屿个数的问题啦。使用一个 DFS 来遍历即可，这个并不难，这道题难就难在需要想出来这种像素化得转化，确实需要灵光一现啊，参见代码如下：
class Solution { public: int regionsBySlashes(vector<string>& grid) { int n = grid.size(), res = 0; vector<vector<int>> nums(3 * n, vector<int>(3 * n)); for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { if (grid[i][j] == '/') { nums[i * 3][j * 3 + 2] = 1; nums[i * 3 + 1][j * 3 + 1] = 1; nums[i * 3 + 2][j * 3] = 1; } else if (grid[i][j] == '\\') { nums[i * 3][j * 3] = 1; nums[i * 3 + 1][j * 3 + 1] = 1; nums[i * 3 + 2][j * 3 + 2] = 1; } } } for (int i = 0; i < nums.size(); ++i) { for (int j = 0; j < nums.size(); ++j) { if (nums[i][j] == 0) { helper(nums, i, j); ++res; } } } return res; } void helper(vector<vector<int>>& nums, int i, int j) { if (i >= 0 && j >= 0 && i < nums.size() && j < nums.size() && nums[i][j] == 0) { nums[i][j] = 1; helper(nums, i - 1, j); helper(nums, i, j + 1); helper(nums, i + 1, j); helper(nums, i, j - 1); } } }; 类似题目：
Github 同步地址:
参考资料：
Metadata
Metadata
Assignees
Labels
No labels