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Description
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��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;
}
};
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