[LeetCode] 531. Lonely Pixel I 孤独的像素 I

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].

给一个只含有黑白像素的图片，找出黑色孤独像素的数量。黑色孤独像素是这个像素所在的行和列都不含有黑色像素。

最基本的想法就是找出每一个黑色像素，然后对相应的行和列进行检查，看是否含有黑色像素。但这种方法肯定含有重复操作，效率肯定不高。

解法：利用数组rows，cols分别记录某行、某列'B'像素的个数。然后遍历一次picture找到符合条件的。

Java:

public int findLonelyPixel(char[][] picture) {
    int n = picture.length, m = picture[0].length;

    int[] rowCount = new int[n], colCount = new int[m];
    for (int i=0;i<n;i++)
        for (int j=0;j<m;j++)
            if (picture[i][j] == 'B') { rowCount[i]++; colCount[j]++; }

    int count = 0;
    for (int i=0;i<n;i++)
        for (int j=0;j<m;j++)
            if (picture[i][j] == 'B' && rowCount[i] == 1 && colCount[j] == 1) count++;

    return count;
}

Java: DFS

public int findLonelyPixel(char[][] picture) {
    int numLone = 0;
    for (int row = 0; row < picture.length; row++) {
        for (int col = 0; col < picture[row].length; col++) {
            if (picture[row][col] == 'W') {
                continue;
            }
            if (dfs(picture, row - 1, col, new int[] {-1, 0}) && dfs(picture, row + 1, col, new int[] {1, 0})
             && dfs(picture, row, col - 1, new int[] {0, -1}) && dfs(picture, row, col + 1, new int[] {0, 1})) {
                numLone++;
            }
        }
    }
    return numLone;
}

// use dfs to find if current pixel is lonely
private boolean dfs(char[][] picture, int row, int col, int[] increase) {
    // base case
    if (row < 0 || row >= picture.length || col < 0 || col >= picture[0].length) {
        return true;
    } else if (picture[row][col] == 'B') {
        return false;
    }
    // recursion
    return dfs(picture, row + increase[0], col + increase[1], increase);
}　　　　

Python:

# Time:  O(m * n)
# Space: O(m + n)
class Solution(object):
    def findLonelyPixel(self, picture):
        """
        :type picture: List[List[str]]
        :rtype: int
        """
        rows, cols = [0] * len(picture),  [0] * len(picture[0])
        for i in xrange(len(picture)):
            for j in xrange(len(picture[0])):
                if picture[i][j] == 'B':
                    rows[i] += 1
                    cols[j] += 1

        result = 0
        for i in xrange(len(picture)):
            if rows[i] == 1:
                for j in xrange(len(picture[0])):
                     result += picture[i][j] == 'B' and cols[j] == 1
        return result

Python:

class Solution(object):
    def findLonelyPixel(self, picture):
        """
        :type picture: List[List[str]]
        :type N: int
        :rtype: int
        """
        return sum(col.count('B') == 1 == picture[col.index('B')].count('B') \
               for col in zip(*picture))

Python:

class Solution(object):
    def findLonelyPixel(self, picture):
        """
        :type picture: List[List[str]]
        :rtype: int
        """
        w, h = len(picture), len(picture[0])
        rows, cols = [0] * w, [0] * h
        for x in range(w):
            for y in range(h):
                if picture[x][y] == 'B':
                    rows[x] += 1
                    cols[y] += 1
        ans = 0
        for x in range(w):
            for y in range(h):
                if picture[x][y] == 'B':
                    if rows[x] == 1:
                        if cols[y] == 1:
                            ans += 1
        return ans　　

Python:

class Solution(object):
  def findLonelyPixel(self, picture):
      """ :type picture: List[List[str]] :rtype: int """
      row = self.find_row(picture)
      column =self.find_colum(picture)

      result = 0
      for x in row:
          for y in column:
              if picture[x][y] == "B":
                  result += 1
      return result

  def find_row(self, picture):
      result = []
      for x in xrange(len(picture)):
          num = 0
          for y in xrange(len(picture[x])):
              if picture[x][y] == "B":
                  num += 1
          if num == 1:
              result.append(x)
      return result

  def find_colum(self, picture):
      result = []
      for y in xrange(len(picture[0])):
          num = 0
          for x in xrange(len(picture)):
              if picture[x][y] == "B":
                  num += 1
          if num == 1:
              result.append(y)
      return result　　

C++:

class Solution {
public:
    int findLonelyPixel(vector<vector<char>>& picture) {
        vector<int> rows = vector<int>(picture.size());
        vector<int> cols = vector<int>(picture[0].size());

        for (int i = 0; i < picture.size(); ++i) {
            for (int j = 0; j < picture[0].size(); ++j) {
                rows[i] += picture[i][j] == 'B';
                cols[j] += picture[i][j] == 'B';
            }
        }

        int result = 0;
        for (int i = 0; i < picture.size(); ++i) {
            if (rows[i] == 1) {
                for (int j = 0; j < picture[0].size() && rows[i] > 0; ++j) {
                    result += picture[i][j] == 'B' && cols[j] == 1;
                }
            }
        }
        return result;
    }
};

C++:

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

　　

　　

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[LeetCode] 533. Lonely Pixel II 孤独的像素 II　　

　

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