LeetCode 531. Lonely Pixel I

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:

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