533. Lonely Pixel II

Given a picture consisting of black and white pixels, and a positive integer N, find the number of black pixels located at some specific row Rand column C that align with all the following rules:

Row R and column C both contain exactly N black pixels.
For all rows that have a black pixel at column C, they should be exactly the same as row R

The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.

Example:

Input:                                            
[['W', 'B', 'W', 'B', 'B', 'W'],    
 ['W', 'B', 'W', 'B', 'B', 'W'],    
 ['W', 'B', 'W', 'B', 'B', 'W'],    
 ['W', 'W', 'B', 'W', 'B', 'W']] 

N = 3
Output: 6
Explanation: All the bold 'B' are the black pixels we need (all 'B's at column 1 and 3).
        0    1    2    3    4    5         column index                                            
0    [['W', 'B', 'W', 'B', 'B', 'W'],    
1     ['W', 'B', 'W', 'B', 'B', 'W'],    
2     ['W', 'B', 'W', 'B', 'B', 'W'],    
3     ['W', 'W', 'B', 'W', 'B', 'W']]    
row index

Take 'B' at row R = 0 and column C = 1 as an example:
Rule 1, row R = 0 and column C = 1 both have exactly N = 3 black pixels. 
Rule 2, the rows have black pixel at column C = 1 are row 0, row 1 and row 2. They are exactly the same as row R = 0.

Note:

The range of width and height of the input 2D array is [1,200].

本题读了好几遍题目也没有怎么读懂，有点小难了，那两个限制条件的大致意思是，第一，某一点为B的点，它的行和列的B的个数都是N，第二个意思是，每一行里面出现的B，B的整个列为B的行必须和该B的行的字符顺序是一样的，代码如下：

 1 public class Solution {
 2     public int findBlackPixel(char[][] picture, int N) {
 3         int row = picture.length;
 4         int col = picture[0].length;
 5         int[] colcount  =new int[col];
 6         Map<String,Integer> map = new HashMap<>();
 7         for(int i=0;i<row;i++){
 8             String s = scanRow(picture,N,colcount,i);
 9             if(s.length()!=0)
10                 map.put(s,map.getOrDefault(s,0)+1);
11         }
12         int res = 0;
13         for(String key:map.keySet()){
14             if(map.get(key)==N){
15                 for(int i=0;i<col;i++){
16                     if(key.charAt(i)=='B'&&colcount[i]==N){
17                         res+=N;
18                     }
19                 }
20             }
21         }
22         return res;
23         
24     }
25     public String scanRow(char[][] picture,int N,int[] colcount,int row){
26         StringBuilder sb = new StringBuilder();
27         int col = picture[0].length;
28         int count = 0;
29         for(int i=0;i<col;i++){
30             if(picture[row][i]=='B'){
31                 count++;
32                 colcount[i]++;
33             }
34             sb.append(picture[row][i]);
35         }
36         if(count==N) return sb.toString();
37         else return "";
38     }
39 }