[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

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

 

LeetCode 题目总结