原题链接在这里：https://leetcode.com/problems/lonely-pixel-i/

题目：

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].

题解：

Have a row array and column array to track how many B on the corresponding row or column.

Iterate the picture for the 1st time and update row and column.

Iterate the prictrue for the 2nd time to accumlate the count when both r[i] and c[j] == 1.

Time Complexity: O(m*n). m = picture.length. n = picture[0].length.

Space: O(m+n).

AC Java:

 1 class Solution {
 2     public int findLonelyPixel(char[][] picture) {
 3         if(picture == null || picture.length == 0 || picture[0].length == 0){
 4             return 0;
 5         }
 6         
 7         int m = picture.length;
 8         int n = picture[0].length;
 9         int [] r = new int[m];
10         int [] c = new int[n];
11         for(int i = 0; i<m; i++){
12             for(int j = 0; j<n; j++){
13                 if(picture[i][j] == 'B'){
14                     r[i]++;
15                     c[j]++;
16                 }
17             }
18         }
19         
20         int res = 0;
21         for(int i = 0; i<m; i++){
22             for(int j = 0; j<n; j++){
23                 if(picture[i][j] == 'B' && r[i] == 1 && c[j] == 1){
24                     res++;
25                 }
26             }
27         }
28         
29         return res;
30     }
31 }