959. Regions Cut By Slashes

2024年6月10日 109次阅读

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 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:

 

 

这题的重点是怎么转化成union find, 

在看了一个题解以后豁然开朗,但是实现的时候又或多或少有点小问题,有的时候index真的是很恼人,

或者说自己没有在脑袋里把它想清楚

 

我们先把一个正方形(小正方形)划分成四个区域,按照对角线划分

 

对于每一个小的正方形, 根据输入,可以有三种merge的方式

1. \:左下方的两个区域进行merge,

2. /:右上方的两个区域merge

3 ‘ ‘: 如果为空, 那么4个区域都需要merge,那么只需要进行三次eg: merge(0, 1), merge(0, 2) merge(0, 3)

 以上就完成了正方形内部的union操作

继续观察,我们可以发现,由于每个小的正方形只有以上三种操作,所以必会得到以下的性质:

1. 正方形的下区域,会跟它下面相邻的正方形的上区域相邻

2. 正方形的右侧区域,会跟它右侧相邻的正方形的左侧相邻, 只要不是最后一行,最后一列,我们都需要进行merge操作

 

 

最后要弄清楚的就是index了.

 

我们吧整个大正方形分成了:

 

N * N 个正方形,然后每个小正方形,所以, 初始状态,一共是 4 * N * N 个集合, 然后通过上述的规则去进行合并,

 

合并的时候,要找到正确的index就可以了

 

 

代码:

 

class Solution {

    
    /**
     * Make each region split into 4 regions,
     * @param grid
     * @return
     */
    public int regionsBySlashes(String[] grid) {

        if(grid == null || grid.length == 0) return 0;

        int row = grid.length;
        int col = grid[0].length();
        int[] parent = new int[row * col * 4];
        // init
        for(int i=0;i<parent.length;i++) parent[i] = i;
        for(int i=0;i<row;i++) {
            char[] line = grid[i].toCharArray();
            for(int j=0;j<col;j++) {
                int index = 4 * i * col + j*4 ;

                if(line[j] == '\\') {
                    union(index+0, index+1, parent);
                     union(index+2, index+3, parent);
                } else if(line[j] == '/') {
                    union(index+0, index+3, parent);
                    union(index+2, index+1, parent);
                } else if(line[j] == ' ') {
                    union(index+0, index+1, parent);
                    union(index+0, index+2, parent);
                    union(index+0, index+3, parent);
                }

                // if it is not the last, row, index merge externally
                if(i<row-1) {
                    union(index+2, index + 4*col+0, parent);
                }
                if(j<col-1) {
                    union(index+1, index + 4 + 3, parent);

                }
            }
        }

        int count=0;
        for(int i=0;i<parent.length;i++) {
            if(parent[i] == i) {
                count++;
            }
        }
        return count;

    }



    private void union(int x, int y, int[] parent) {
        int xp = find(x, parent);
        int yp = find(y, parent);
        if(xp != yp) {
            parent[xp] = yp;
        }
    }

    private int find(int x, int[] parent) {
        if(parent[x] != x) {
            return parent[x] = find(parent[x], parent);
        }
        return x;
    }
}

 

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