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].
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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