Question

531. 孤独像素 I

English Version

题目描述

给你一个大小为 m x n 的图像 picture ，图像由黑白像素组成，'B' 表示黑色像素，'W' 表示白色像素，请你统计并返回图像中 黑色 孤独像素的数量。

黑色孤独像素 的定义为：如果黑色像素 'B' 所在的同一行和同一列不存在其他黑色像素，那么这个黑色像素就是黑色孤独像素。

 

示例 1：

输入：picture = [["W","W","B"],["W","B","W"],["B","W","W"]]
输出：3
解释：全部三个 'B' 都是黑色的孤独像素

示例 2：

输入：picture = [["B","B","B"],["B","B","W"],["B","B","B"]]
输出：0

 

提示：

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

解法

方法一

class Solution:
  def findLonelyPixel(self, picture: List[List[str]]) -> int:
    m, n = len(picture), len(picture[0])
    rows, cols = [0] * m, [0] * n
    for i in range(m):
      for j in range(n):
        if picture[i][j] == 'B':
          rows[i] += 1
          cols[j] += 1
    res = 0
    for i in range(m):
      if rows[i] == 1:
        for j in range(n):
          if picture[i][j] == 'B' and cols[j] == 1:
            res += 1
            break
    return res

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

class Solution {
public:
  int findLonelyPixel(vector<vector<char>>& picture) {
    int m = picture.size(), n = picture[0].size();
    vector<int> rows(m);
    vector<int> cols(n);
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (picture[i][j] == 'B') {
          ++rows[i];
          ++cols[j];
        }
      }
    }
    int res = 0;
    for (int i = 0; i < m; ++i) {
      if (rows[i] == 1) {
        for (int j = 0; j < n; ++j) {
          if (picture[i][j] == 'B' && cols[j] == 1) {
            ++res;
            break;
          }
        }
      }
    }
    return res;
  }
};

func findLonelyPixel(picture [][]byte) int {
  m, n := len(picture), len(picture[0])
  rows := make([]int, m)
  cols := make([]int, n)
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      if picture[i][j] == 'B' {
        rows[i]++
        cols[j]++
      }
    }
  }
  res := 0
  for i := 0; i < m; i++ {
    if rows[i] == 1 {
      for j := 0; j < n; j++ {
        if picture[i][j] == 'B' && cols[j] == 1 {
          res++
          break
        }
      }
    }
  }
  return res
}