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].
这道题是之前那道Lonely Pixel I的拓展，我开始以为这次要考虑到对角线的情况，可是这次题目却完全换了一种玩法。给了一个整数N，说对于均含有N个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。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，参见代码如下：
解法一：
public: int findBlackPixel(vector<vector<char>>& picture, int N) { if (picture.empty() || picture[0].empty()) return 0; int m = picture.size(), n = picture[0].size(), res = 0, k = 0; vector<int> rowCnt(m, 0), colCnt(n, 0); vector<string> rows(m, ""); for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { rows[i].push_back(picture[i][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 (rowCnt[i] == N && colCnt[j] == N) { for (k = 0; k < m; ++k) { if (picture[k][j] == 'B') { if (rows[i] != rows[k]) break; } } if (k == m) { res += colCnt[j]; colCnt[j] = 0; } } } } return res; } };
看到论坛中的比较流行的解法是用哈希表来做的，建立黑像素出现个数为N的行和其出现次数之间的映射，然后我们就只需要统计每列的黑像素的个数，然后我们遍历哈希表，找到出现次数刚好为N的行，说明矩阵中有N个相同的该行，而且该行中的黑像素的个数也刚好为N个，那么第二个条件就已经满足了，我们只要再满足第一个条件就行了，我们在找黑像素为N个的列就行了，有几列就加几个N即可，参见代码如下：
解法二：
public: int findBlackPixel(vector<vector<char>>& picture, int N) { if (picture.empty() || picture[0].empty()) return 0; int m = picture.size(), n = picture[0].size(), res = 0; vector<int> colCnt(n, 0); unordered_map<string, int> u; for (int i = 0; i < m; ++i) { int cnt = 0; for (int j = 0; j < n; ++j) { if (picture[i][j] == 'B') { ++colCnt[j]; ++cnt; } } if (cnt == N) ++u[string(picture[i].begin(), picture[i].end())]; } for (auto a : u) { if (a.second != N) continue; for (int i = 0; i < n; ++i) { res += (a.first[i] == 'B' && colCnt[i] == N) ? N : 0; } } return res; } };
参考资料：
https://discuss.leetcode.com/topic/81686/verbose-java-o-m-n-solution-hashmap/2
https://discuss.leetcode.com/topic/87164/a-c-solution-based-on-the-top-rated-issue
本文转自博客园Grandyang的博客，原文链接：[LeetCode] Lonely Pixel II 孤独的像素之二
，如需转载请自行联系原博主。
