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:

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

本题有点类似于皇后问题，思路是，第一遍找出有B的字符，然后把它的行和列分别+1，第二遍遍历的时候，找出有B的字符并且，行和列都是1的字符，count++；

代码如下：

public class Solution {
    public int findLonelyPixel(char[][] picture) {
        int count = 0;
        int row = picture.length;
        int col = picture[0].length;
        int[] rows = new int[row];
        int[] cols = new int[col];
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(picture[i][j]=='B'){
                    rows[i]++;
                    cols[j]++;
                }
            }
        }
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(picture[i][j]=='B'&&rows[i]==1&&cols[j]==1){
                    count++;
                }
            }
        }
        return count;
    }
}