[LeetCode] 531. Lonely Pixel I

Given an m x n picture consisting of black 'B' and white 'W' pixels, return the number of black lonely pixels.

A black lonely pixel is a character 'B' that located at a specific position where the same row and same column don't have any other black pixels.

Example 1:

Input: picture = [["W","W","B"],["W","B","W"],["B","W","W"]]
Output: 3
Explanation: All the three 'B's are black lonely pixels.

Example 2:

Input: picture = [["B","B","B"],["B","B","W"],["B","B","B"]]
Output: 0

Constraints:

• m == picture.length
• n == picture[i].length
• 1 <= m, n <= 500
• picture[i][j] is 'W' or 'B'.

孤独像素。

这道题不难，我直接介绍思路。

首先我用两个数组分别记录 input 矩阵中每一行和每一列出现了多少个字母 B。之后我再次遍历 input 矩阵，当我再次遇到字母 B 的时候，我检查一下当前这个 B 是不是他所在行和所在列上唯一的一个 B。

时间O(mn)

空间O(m + n)

Java实现

 1 class Solution {
 2     public int findLonelyPixel(char[][] picture) {
 3         int m = picture.length;
 4         int n = picture[0].length;
 5         int[] rows = new int[m];
 6         int[] cols = new int[n];
 7         for (int i = 0; i < picture.length; i++) {
 8             for (int j = 0; j < picture[0].length; j++) {
 9                 if (picture[i][j] == 'B') {
10                     rows[i]++;
11                     cols[j]++;
12                 }
13             }
14         }
15         
16         int count = 0;
17         for (int i = 0; i < m; i++) {
18             for (int j = 0; j < n; j++) {
19                 if (picture[i][j] == 'B' && rows[i] == 1 && cols[j] == 1) {
20                     count++;
21                 }
22             }
23         }
24         return count;
25     }
26 }

LeetCode 题目总结