[Leetcode] 533. Lonely Pixel II 解题报告

题目：

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:

1. Row R and column C both contain exactly N black pixels.
2. 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:

1. The range of width and height of the input 2D array is [1,200].

思路：

和上一道题目类似，只不过加了一个额外条件，就是在某个位置上的黑点如果要符合条件，在该列上为黑点的所有行中的内容还需要相等。为此我们定义了一个哈希表和一个string构成的行向量，这是因为将vector转化为string之后，将便于比较异同；另外用哈希表可以将行向量string映射成为其出现的次数。这样后面的比较就更加方便了。此时picture[r][c]如果满足如下几条则为符合条件的点：1）row_pixels[r] == N；2）col_pixels[c] == N；3）same_rows[rows[r]] == N。

代码：

class Solution {
public:
    int findBlackPixel(vector>& picture, int N) {
        if (picture.size() == 0 || picture[0].size() == 0) {
            return 0;
        }
        vector row_pixels(picture.size(), 0);
        vector col_pixels(picture[0].size(), 0);
        unordered_map same_rows;       // map the row string to appearance count
        vector rows;      // convert the vector to string for easier comparison
        for (int r = 0; r < picture.size(); ++r) {
            string s;
            for (int c = 0; c < picture[r].size(); ++c) {
                if (picture[r][c] == 'B') {
                    ++row_pixels[r], ++col_pixels[c];
                }
                s.push_back(picture[r][c]);
            }
            ++same_rows[s];
            rows.push_back(s);
        }
        int ret = 0;
        for (int r = 0; r < picture.size(); ++r) {
            if (row_pixels[r] == N && same_rows[rows[r]] == N) {
                for (int c = 0; c < picture[r].size(); ++c) {
                    if (picture[r][c] == 'B' && col_pixels[c] == N) {
                        ++ret;
                    }
                }
            }
        }
        return ret;
    }
};