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].
题目标签：Array
题目给了我们一个2D array picture 和一个N，让我们从picture 中找到 符合 两条规则的black pixel的个数。
rule 2 真是一开始没理解，琢磨了一会，去看了解释才明白。
rule 2 说的是，当满足了rule 1 有一个在row R 和 column C 的black pixel，这个像素的行 和列 都要有N个 black pixels之后，
还需要满足column C 中 有black pixels 的 rows 都要和 R 这一行row 一摸一样。
建立一个Map，把row String（把每一行char组成string） 当作key， 把这一个row 出现过的次数 当作value；还需要建立一个cols array，来记录每一列的black pixels个数。
遍历picture：
1. 记录每一列的B 个数；
2. 记录每一行的B 个数，只有等于N的情况下，才把row string 存入map。（每一行的black pixels个数在这里就被完成了，所以不需要额外空间来存放）
遍历map 的keySet（有N个B的行）：
1. 如果这个row出现的次数 不等于 N的话， 说明 不满足rule 1的一列里要有N个B的条件。 因为如果当前 row 出现了N次，而且row 在之前已经满足了一行里有N个B的条件。所以每行里肯定有B，然后有B的一列里也会有N个B。
当满足了row出现的次数 等于N 之后，意味着也满足了rule 2，因为这N个rows 都是一摸一样的。
2. 当满足了上面这个条件后，遍历所有列：把每列的B的个数N加入res。
Java Solution:
Runtime beats 71.04%
完成日期：09/25/2017
关键词：Array, HashMap
关键点：建立一个HashMap，使得每一行的string 和 它的出现次数形成映射（满足rule1 rule2）；建立array cols 来辅助找到每一列中符合标准的B的数量
class Solution
{
public int findBlackPixel(char[][] picture, int N)
{
int m = picture.length;
int n = picture[0].length;
int[] cols = new int[n];
HashMap<String, Integer> map = new HashMap<>();
int res = 0; // iterate picture
for(int i=0; i<m; i++) // rows
{
int count = 0;
StringBuilder sb = new StringBuilder(); for(int j=0; j<n; j++) // cols
{
if(picture[i][j] == 'B')
{
cols[j]++;
count++;
}
sb.append(picture[i][j]);
} if(count == N) // only store rowString into map when this row has N B
{
String curRow = sb.toString();
map.put(curRow, map.getOrDefault(curRow, 0) + 1); // count how many rows are same
} } // iterate keySet
for(String row : map.keySet())
{
if(map.get(row) != N) // if value is not N, meaning rule 1 (columns need to have N Bs) is not satisfied.
continue;
// if value is N, meaning rule 2 is also satisfied because all these rows are same.
for(int j=0; j<n; j++) // iterate column
{ // count Bs in this column
if(row.charAt(j) == 'B' && cols[j] == N)
res += N;
}
} return res;
}
}
参考资料：
https://discuss.leetcode.com/topic/81686/verbose-java-o-m-n-solution-hashmap
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