[LeetCode] Lonely Pixel II 孤獨的像素之二

 

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].

 

這道題是之前那道Lonely Pixel I的拓展，我開始以為這次要考慮到對角線的情況，可是這次題目卻完全換了一種玩法。給了一個整數N，說對於均含有N個個黑像素的某行某列，如果該列中所有的黑像素所在的行都相同的話，該列的所有黑像素均為孤獨的像素，讓我們統計所有的這樣的孤獨的像素的個數。那么跟之前那題類似，我們還是要統計每一行每一列的黑像素的個數，而且由於條件二中要比較各行之間是否相等，如果一個字符一個字符的比較寫起來比較麻煩，我們可以用個trick，把每行的字符連起來，形成一個字符串，然后直接比較兩個字符串是否相等會簡單很多。然后我們遍歷每一行和每一列，如果某行和某列的黑像素剛好均為N，我們遍歷該列的所有黑像素，如果其所在行均相等，則說明該列的所有黑像素均為孤獨的像素，將個數加入結果res中，然后將該行的黑像素統計個數清零，以免重復運算，這樣我們就可以求出所有的孤獨的像素了，參見代碼如下：

 

解法一：

class Solution {
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;
    }
};

 

看到論壇中的比較流行的解法是用哈希表來做的，建立黑像素出現個數為N的行和其出現次數之間的映射，然后我們就只需要統計每列的黑像素的個數，然后我們遍歷哈希表，找到出現次數剛好為N的行，說明矩陣中有N個相同的該行，而且該行中的黑像素的個數也剛好為N個，那么第二個條件就已經滿足了，我們只要再滿足第一個條件就行了，我們在找黑像素為N個的列就行了，有幾列就加幾個N即可，參見代碼如下：

 

解法二：

class Solution {
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;
    }
};

 

類似題目：

Lonely Pixel I

 

參考資料：

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

 

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