Leetcode解题: Lonely Pixel (531)

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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:

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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].

解法1：O(MN) Time + O(M + N) Space

用两个数组分别记录每一行， 每一列的pixel的个数，然后遍历一遍矩阵，对于每一个B的位置查看当前行列的B的个数。
Java

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public class Solution {
    public int findLonelyPixel(char[][] picture) {
        if (picture.length == 0 || picture[0].length == 0) {
            return 0;
        }
        
        int n = picture.length;
        int m = picture[0].length;
        int[] row = new int[n];
        int[] col = new int[m];
        
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                if (picture[i][j] == 'B') {
                    row[i]++;
                    col[j]++;
                }
            }
        }
        
        int count = 0;
        for (int i = 0; i < n;++i) {
            for (int j = 0; j < m; ++j) {
                if (picture[i][j] == 'B') {
                    if (row[i] == 1 && col[j] == 1) {
                        ++count;
                    }
                }
            }
        }
        
        return count;
        
    }
}