531. Lonely Pixel I

• 44.4%

https://leetcode.com/contest/leetcode-weekly-contest-22/problems/lonely-pixel-i/

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:

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

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cpp

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my code:

与第200道，小岛的题目很类似，可以考虑参考。但是有不一样，这里建立两个数组，保证横纵坐标都是1，且此处为‘B’就是要找的点，计数。

class Solution {
public:
    int findLonelyPixel(vector<vector<char>>& picture) {
        // if(picture==NULL) return 0;
        if(picture.size()==0) return 0;
        if(picture[0].size()==0) return 0;
        int m = picture.size();
        int n = picture[0].size();
        vector<int> dp1(m, 0);
        vector<int> dp2(n, 0);
        for(int i=0; i<m; i++){
            int tmp = 0;
            for(int j=0; j<n; j++){
                if(picture[i][j]=='B')
                    tmp++;
            }
            dp1[i] = tmp;
        }
        
        for(int i=0; i<n; i++){
            int tmp = 0;
            for(int j=0; j<m; j++){
                if(picture[j][i]=='B')
                    tmp++;
            }
            dp2[i] = tmp;
        }
        
        int res = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(dp1[i]==1 && dp2[j]==1 && picture[i][j]=='B')
                    res++;
            }
        }
        return res;
    }
};