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：

 1 public class Solution {
 2     public int findLonelyPixel(char[][] picture) {
 3         int count = 0;
 4         int row = picture.length;
 5         int col = picture[0].length;
 6         int[] rows = new int[row];
 7         int[] cols = new int[col];
 8         for(int i=0;i<row;i++){
 9             for(int j=0;j<col;j++){
10                 if(picture[i][j]==‘B‘){
11                     rows[i]++;
12                     cols[j]++;
13                 }
14             }
15         }
16         for(int i=0;i<row;i++){
17             for(int j=0;j<col;j++){
18                 if(picture[i][j]==‘B‘&&rows[i]==1&&cols[j]==1){
19                     count++;
20                 }
21             }
22         }
23         return count;
24     }
25 }