Given a picture consisting of black and white pixels, and a positive integer N, find the number of black pixels located at some specific row R and column C that align with all the following rules:
Row R and column C both contain exactly N black pixels.
For all rows that have a black pixel at column C, they should be exactly the same as row R
The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.

Example:
Input:                                            
[['W', 'B', 'W', 'B', 'B', 'W'],    
 ['W', 'B', 'W', 'B', 'B', 'W'],    
 ['W', 'B', 'W', 'B', 'B', 'W'],    
 ['W', 'W', 'B', 'W', 'B', 'W']] 

N = 3
Output: 6
Explanation: All the bold 'B' are the black pixels we need (all 'B's at column 1 and 3).
        0    1    2    3    4    5         column index                                            
0    [['W', 'B', 'W', 'B', 'B', 'W'],    
1     ['W', 'B', 'W', 'B', 'B', 'W'],    
2     ['W', 'B', 'W', 'B', 'B', 'W'],    
3     ['W', 'W', 'B', 'W', 'B', 'W']]    
row index

Take 'B' at row R = 0 and column C = 1 as an example:
Rule 1, row R = 0 and column C = 1 both have exactly N = 3 black pixels. 
Rule 2, the rows have black pixel at column C = 1 are row 0, row 1 and row 2. They are exactly the same as row R = 0.

Note:
The range of width and height of the input 2D array is [1,200].

Solution：

思路：
給了一個(gè)整數(shù)N，說(shuō)對(duì)于均含有N個(gè)黑像素的某行某列讲冠，如果該列中所有的黑像素所在的行都相同的話椅寺，該列的所有黑像素均為孤獨(dú)的像素浆西，讓我們統(tǒng)計(jì)所有的這樣的孤獨(dú)的像素的個(gè)數(shù)歧蕉。

那么跟之前那題類似汰蓉，我們還是要統(tǒng)計(jì)每一行每一列的黑像素的個(gè)數(shù)，而且由于條件二中要比較各行之間是否相等奋姿，如果一個(gè)字符一個(gè)字符的比較寫(xiě)起來(lái)比較麻煩锄开，我們可以用個(gè)trick，把每行的字符連起來(lái)称诗，形成一個(gè)字符串萍悴，然后直接比較兩個(gè)字符串是否相等會(huì)簡(jiǎn)單很多(用hashmap(string, count))。然后我們遍歷每一行和每一列寓免，如果某行和某列的黑像素剛好均為N癣诱，我們遍歷該列的所有黑像素，如果其所在行均相等袜香，則說(shuō)明該列的所有黑像素均為孤獨(dú)的像素撕予，將個(gè)數(shù)加入結(jié)果res中，然后將該行的黑像素統(tǒng)計(jì)個(gè)數(shù)清零蜈首，以免重復(fù)運(yùn)算实抡，這樣我們就可以求出所有的孤獨(dú)的像素了，

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

// 孤獨(dú)的點(diǎn)要滿足:
// (1) 該行有N個(gè)黑點(diǎn)
// (2) 該列有N個(gè)黑點(diǎn)
// (3) 該列中所有的黑像素所在的行都相同
public class Solution {
    public int findBlackPixel(char[][] picture, int N) {
        if(picture == null || picture.length == 0 || picture[0].length == 0) return 0;
        int m = picture.length;
        int n = picture[0].length;
        
        Map<String, Integer> map = new HashMap<>();
        int[] colCount = new int[n];
        
        // (預(yù)備3)行編碼存入map欢策，(預(yù)備2)統(tǒng)計(jì)列的B吆寨，并check(1)該行B的個(gè)數(shù)是否為N
        for (int i = 0; i < m; i++) {
            String key = scanRow(picture, i, N, colCount);
            if (key.length() != 0) {
                map.put(key, map.getOrDefault(key, 0) + 1);
            }
        }
        
        // check(3) 及 check(2)該列B的個(gè)數(shù)是否為N
        int result = 0;
        for (String key : map.keySet()) {
            if (map.get(key) == N) {
                for (int j = 0; j < n; j++) {
                    if (key.charAt(j) == 'B' && colCount[j] == N) {
                        result += N;
                    }
                }
            }
        }
        
        return result;
    }
    
    private String scanRow(char[][] picture, int row, int target, int[] colCount) {
        int n = picture[0].length;
        int rowCount = 0;
        StringBuilder sb = new StringBuilder();
        
        for (int j = 0; j < n; j++) {
            if (picture[row][j] == 'B') {
                rowCount++;
                colCount[j]++;
            }
            sb.append(picture[row][j]);
        }
        
        if (rowCount == target) return sb.toString();
        return "";
    }
}