531. Lonely Pixel I

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].

Solution1：

思路：
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
(也可以在原数组上存，减少Space)
Time Complexity: O(mn) Space Complexity: O(m+n)

https://leetcode.com/problems/lonely-pixel-i/discuss/100018/

Solution1 Code：

class Solution {
    public int findLonelyPixel(char[][] picture) {
        int n = picture.length, m = picture[0].length;

        int[] rowCount = new int[n], colCount = new int[m];
        for (int i=0;i<n;i++) {
            for (int j=0;j<m;j++) { 
                if (picture[i][j] == 'B') { 
                    rowCount[i]++; colCount[j]++; 
                }
            }
        }

        int count = 0;
        for (int i=0;i<n;i++) {
            for (int j=0;j<m;j++) { 
                if (picture[i][j] == 'B' && rowCount[i] == 1 && colCount[j] == 1) count++;
            }
        }

        return count;
    }

}