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:
- 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：
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; } };
本文转自博客园Grandyang的博客，原文链接：[LeetCode] Lonely Pixel I 孤独的像素之一
，如需转载请自行联系原博主。
