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;
     }
 }