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:
https://github.com/grandyang/leetcode/issues/959
参考资料：
https://leetcode.com/problems/regions-cut-by-slashes/
LeetCode All in One 题目讲解汇总(持续更新中…)
转载请注明来源于 Grandyang 的博客 (grandyang.com)，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 grandyang@qq.com
