959. Regions Cut By Slashes
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 ais 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 becausecharacters are escaped, "\/" refers to /, and "/\" refers to /.)
The 2x2 grid is as follows:
Example 4:
Input:
[
"/\",
"\/"
]
Output: 5
Explanation: (Recall that becausecharacters 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 <= 30
gridi is either '/', '', or ' '.
難度：medium
題目：
給定一個由1X1的方塊組成的網格，每個方塊由/‘ ’（空字元） 3 個元素組成，這些元素將大方塊分成若干小的連續區域。求連續區域的個數。
思路：
將每個小方格用3 Ｘ 3的0、1陣列表示。因此題目就轉成了0、1矩陣中求0的連續區域個數。
/ 轉換成
001
010
100
轉換成
100
010
001
‘ ’ 轉換成
000
000
000
Runtime: 12 ms, faster than 85.37% of Java online submissions for Regions Cut By Slashes.
class Solution {
/**
*001
* / -> 010
*100
*
*100
* \ -> 010
*001
*
*000
* ' '->000
*000
*
* convert to find the isolated 0s by 1
*/
public int regionsBySlashes(String[] grid) {
int row = grid.length;
int trippleRow = 3 * row;
int[][] iGrid = new int[trippleRow][trippleRow];
for (int i = 0; i < trippleRow; i += 3) {
char[] sc = grid[i / 3].toCharArray();
for (int j = 0; j < trippleRow; j += 3) {
char c = sc[j / 3];
if ('/' == c) {
iGrid[i][j + 2] = iGrid[i + 1][j + 1] = iGrid[i + 2][j] = 1;
} else if ('\\' == c) {
iGrid[i][j] = iGrid[i + 1][j + 1] = iGrid[i + 2][j + 2] = 1;
}
}
}
int count = 0;
for (int i = 0; i < trippleRow; i++) {
for (int j = 0; j < trippleRow; j++) {
if (0 == iGrid[i][j]) {
colorGrid(iGrid, i, j);
count++;
}
}
}
return count;
}
private void colorGrid(int[][] grid, int i, int j) {
grid[i][j] = 1;
// up
if (0 == grid[Math.max(i - 1, 0)][j]) {
colorGrid(grid, i - 1, j);
}
// down
if (0 == grid[Math.min(i + 1, grid.length - 1)][j]) {
colorGrid(grid, i + 1, j);
}
// left
if (0 == grid[i][Math.max(j - 1, 0)]) {
colorGrid(grid, i, j - 1);
}
// right
if (0 == grid[i][Math.min(j + 1, grid[0].length - 1)] ) {
colorGrid(grid, i, j + 1);
}
}
}