• [LeetCode] Lonely Pixel I 孤独的像素之一


    Given a picture consisting of black and white pixels, find the number of black lonely pixels.

    The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.

    A black lonely pixel is character 'B' that located at a specific position where the same row and same column don't have any other black pixels.

    Example:

    Input: 
    [['W', 'W', 'B'],
     ['W', 'B', 'W'],
     ['B', 'W', 'W']]
    
    Output: 3
    Explanation: All the three 'B's are black lonely pixels.
    

    Note:

    1. The range of width and height of the input 2D array is [1,500].

    这道题定义了一种孤独的黑像素,就是该黑像素所在的行和列中没有其他的黑像素,让我们找出所有的这样的像素。那么既然对于每个黑像素都需要查找其所在的行和列,为了避免重复查找,我们可以统一的扫描一次,将各行各列的黑像素的个数都统计出来,然后再扫描所有的黑像素一次,看其是否是该行该列唯一的存在,是的话就累加计数器即可,参见代码如下:

    class Solution {
    public:
        int findLonelyPixel(vector<vector<char>>& picture) {
            if (picture.empty() || picture[0].empty()) return 0;
            int m = picture.size(), n = picture[0].size(), res = 0;
            vector<int> rowCnt(m, 0), colCnt(n, 0);
            for (int i = 0; i < m; ++i) {
                for (int j = 0; j < n; ++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 (picture[i][j] == 'B') {
                        if (rowCnt[i] == 1 && colCnt[j] == 1) {
                            ++res;
                        }
                    }
                }
            }
            return res;
        }
    };

    LeetCode All in One 题目讲解汇总(持续更新中...)

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  • 原文地址:https://www.cnblogs.com/grandyang/p/6754499.html
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