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].
题目标签：Array
题目给了我们一个2d picture array，让我们找出有几个孤独的像素B。孤独的像素B的行和列中只有自己一个像素。
可以建立2个 array， 分别是 row 和 col， 它们的size 就是picture里的行和列的size。
第一次遍历picture：如果是B，就把B的row 和 column 在 row array 和 col array 里对应位置 加1。目的是为了记录每一行和每一列里有多少个B。
第二次遍历picture：如果是B，就到对应的 row 和 col array 里， 如果 row 和 col 的值都是1 的话，说明这一个B 是 row 这一行里， 和 col 这一列里唯一的B。累计计数到res。
Java Solution:
Runtime beats 80.41%
完成日期：09/24/2017
关键词：Array
关键点：设立row array & col array 记录每一行每一列里有多少个B；孤单像素所在的行和列只有它自己
1 class Solution
2 {
3 public int findLonelyPixel(char[][] picture)
4 {
5 int n = picture.length;
6 int m = picture[0].length;
7 // create two array for counting B
8 int[] row = new int[n];
9 int[] col = new int[m];
10
11 int res = 0; // counts longly black
12
13 // 1st iteration, if find a B, increase its row and col number in two array
14 for(int i=0; i<n; i++)
15 {
16 for(int j=0; j<m; j++)
17 {
18 if(picture[i][j] == ‘B‘)
19 {
20 row[i]++;
21 col[j]++;
22 }
23 }
24 }
25
26
27 // 2nd iteration, if find a B, check its row and col are both 1, meaning lonly B
28 for(int i=0; i<n; i++)
29 for(int j=0; j<m; j++)
30 if(picture[i][j] == ‘B‘ && row[i] == 1 && col[j] == 1)
31 res++;
32
33 return res;
34 }
35 }
参考资料：
https://discuss.leetcode.com/topic/81680/java-o-nm-time-with-o-n-m-space-and-o-1-space-solutions
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