Given a picture consisting of black and white pixels, find the number of black lonely pixels.

The picture is represented by a 2D char array consisting of ‘B’ and ‘W’, which means black and white pixels respectively.

A black lonely pixel is character ‘B’ that located at a specific position where the same row and same column don’t have any other black pixels.

Example:

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

Output: 3
Explanation: All the three 'B's are black lonely pixels.

Note:

1. The range of width and height of the input 2D array is [1,500].

这道题定义了一种孤独的黑像素，就是该黑像素所在的行和列中没有其他的黑像素，让我们找出所有的这样的像素。那么既然对于每个黑像素都需要查找其所在的行和列，为了避免重复查找，我们可以统一的扫描一次，将各行各列的黑像素的个数都统计出来，然后再扫描所有的黑像素一次，看其是否是该行该列唯一的存在，是的话就累加计数器即可，参见代码如下：

class Solution {
public:
    int findLonelyPixel(vector<vector<char>>& picture) {
        if (picture.empty() || picture[0].empty()) return 0;
        int m = picture.size(), n = picture[0].size(), res = 0;
        vector<int> rowCnt(m, 0), colCnt(n, 0);
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (picture[i][j] == 'B') {
                    ++rowCnt[i];
                    ++colCnt[j];
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (picture[i][j] == 'B') {
                    if (rowCnt[i] == 1 && colCnt[j] == 1) {
                        ++res;
                    }
                }
            }
        }
        return res;
    }
};