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:

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;

     }

 }

巴特西

531. Lonely Pixel I

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