Given a picture consisting of black and white pixels, and a positive integer N, find the number of black pixels located at some specific row R and column C that align with all the following rules:
Row R and column C both contain exactly N black pixels.
For all rows that have a black pixel at column C, they should be exactly the same as row R
The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.
Example:
Input:
[['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'W', 'B', 'W', 'B', 'W']]
N = 3
Output: 6
Explanation: All the bold 'B' are the black pixels we need (all 'B's at column 1 and 3).
0 1 2 3 4 5 column index
0 [['W', 'B', 'W', 'B', 'B', 'W'],
1 ['W', 'B', 'W', 'B', 'B', 'W'],
2 ['W', 'B', 'W', 'B', 'B', 'W'],
3 ['W', 'W', 'B', 'W', 'B', 'W']]
row index
Take 'B' at row R = 0 and column C = 1 as an example:
Rule 1, row R = 0 and column C = 1 both have exactly N = 3 black pixels.
Rule 2, the rows have black pixel at column C = 1 are row 0, row 1 and row 2. They are exactly the same as row R = 0.
Note:
The range of width and height of the input 2D array is [1,200].
Solution:
思路:
给了一个整数N,说对于均含有N个黑像素的某行某列,如果该列中所有的黑像素所在的行都相同的话,该列的所有黑像素均为孤独的像素,让我们统计所有的这样的孤独的像素的个数。
那么跟之前那题类似,我们还是要统计每一行每一列的黑像素的个数,而且由于条件二中要比较各行之间是否相等,如果一个字符一个字符的比较写起来比较麻烦,我们可以用个trick,把每行的字符连起来,形成一个字符串,然后直接比较两个字符串是否相等会简单很多(用hashmap(string, count))。然后我们遍历每一行和每一列,如果某行和某列的黑像素刚好均为N,我们遍历该列的所有黑像素,如果其所在行均相等,则说明该列的所有黑像素均为孤独的像素,将个数加入结果res中,然后将该行的黑像素统计个数清零,以免重复运算,这样我们就可以求出所有的孤独的像素了,
Time Complexity: O(N) Space Complexity: O(N)
Solution Code:
// 孤独的点要满足:
// (1) 该行有N个黑点
// (2) 该列有N个黑点
// (3) 该列中所有的黑像素所在的行都相同
public class Solution {
public int findBlackPixel(char[][] picture, int N) {
if(picture == null || picture.length == 0 || picture[0].length == 0) return 0;
int m = picture.length;
int n = picture[0].length;
Map<String, Integer> map = new HashMap<>();
int[] colCount = new int[n];
// (预备3)行编码存入map,(预备2)统计列的B,并check(1)该行B的个数是否为N
for (int i = 0; i < m; i++) {
String key = scanRow(picture, i, N, colCount);
if (key.length() != 0) {
map.put(key, map.getOrDefault(key, 0) + 1);
}
}
// check(3) 及 check(2)该列B的个数是否为N
int result = 0;
for (String key : map.keySet()) {
if (map.get(key) == N) {
for (int j = 0; j < n; j++) {
if (key.charAt(j) == 'B' && colCount[j] == N) {
result += N;
}
}
}
}
return result;
}
private String scanRow(char[][] picture, int row, int target, int[] colCount) {
int n = picture[0].length;
int rowCount = 0;
StringBuilder sb = new StringBuilder();
for (int j = 0; j < n; j++) {
if (picture[row][j] == 'B') {
rowCount++;
colCount[j]++;
}
sb.append(picture[row][j]);
}
if (rowCount == target) return sb.toString();
return "";
}
}
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