java 解法, 执行用时: 2 ms, 内存消耗: 43 MB, 提交时间: 2023-10-22 00:00:49

class Solution {
    public int findLonelyPixel(char[][] picture) {
        int ans = 0;
        int rowN = picture.length;
        int colN = picture[0].length;
        int[] row = new int[rowN];
        int[] col = new int[colN];
        List<Integer[]> list = new ArrayList<>();
        for (int i = 0; i < rowN; i++) {
            for (int j = 0; j < colN; j++) {
                if (picture[i][j] == 'B') {
                    row[i]++;
                    col[j]++;
                    list.add(new Integer[]{i, j});
                }
            }
        }
        for (Integer[] blackPoint : list) {
            int i = blackPoint[0];
            int j = blackPoint[1];
            if (row[i] == 1 && row[i] == col[j]) {
                ans++;
            }
        }
        return ans;
    }
}

python3 解法, 执行用时: 76 ms, 内存消耗: 21.3 MB, 提交时间: 2023-10-22 00:00:18

class Solution:
    def findLonelyPixel2(self, picture: List[List[str]]) -> int:
        row, col = [collections.Counter(r)['B'] for r in picture], [collections.Counter(c)['B'] for c in zip(*picture)]
        return sum((row[r] == 1 and col[c] == 1 and picture[r][c] == 'B') for r in range(len(picture)) for c in range(len(picture[0])))
        
    def findLonelyPixel(self, picture: List[List[str]]) -> int:
        R, C = len(picture), len(picture[0])
        row_sum = [0 for _ in range(R)]
        col_sum = [0 for _ in range(C)]
        for r in range(R):
            for c in range(C):
                if picture[r][c] == 'B':
                    row_sum[r] += 1
                    col_sum[c] += 1
        res = 0
        for r in range(R):
            if row_sum[r] == 1:
                for c in range(C):
                    if picture[r][c] == 'B' and col_sum[c] == 1:
                        res += 1
        return res

golang 解法, 执行用时: 4 ms, 内存消耗: 5.7 MB, 提交时间: 2023-10-21 23:59:30

func findLonelyPixel(picture [][]byte) int {
    x, y := map[int]int{}, map[int]int{}
    for i := 0; i < len(picture); i++ {
        for j := 0; j < len(picture[i]); j++ {
            if picture[i][j] == 'B' {
                x[i]++
                y[j]++
            }
        }
    }
    ret := 0
    for i := 0; i < len(picture); i++ {
        for j := 0; j < len(picture[i]); j++ {
            if picture[i][j] == 'B' && x[i] == 1 && y[j] == 1 {
                ret++
            }
        }
    } 
    return ret
}

cpp 解法, 执行用时: 72 ms, 内存消耗: 16.5 MB, 提交时间: 2023-10-21 23:59:12

class Solution {
public:
    int findLonelyPixel(vector<vector<char>>& picture) {
        int n=(int)picture.size(),m=(int)picture[0].size();
        int ans=0;
        for (int i=0;i<n;++i){
            for (int j=0;j<m;++j)if(picture[i][j]=='B'){
                int col=0,row=0;
                for (int k=0;k<m;++k){
                    row+=picture[i][k]=='B';
                }
                for (int k=0;k<n;++k){
                    col+=picture[k][j]=='B';
                }
                if (row==1 && col==1) ans++;
            }
        }
        return ans;
    }
};

class Solution2 {
    int row[505],col[505];
public:
    int findLonelyPixel(vector<vector<char>>& picture) {
        int n=(int)picture.size(),m=(int)picture[0].size();
        for (int i=0;i<n;++i){
            for (int j=0;j<m;++j){
                row[i]+=picture[i][j]=='B';
                col[j]+=picture[i][j]=='B';
            }
        }
        int ans=0;
        for (int i=0;i<n;++i)if(row[i]==1){
            for (int j=0;j<m;++j)if(picture[i][j]=='B'){
                ans+=col[j]==1;
            }
        }
        return ans;
    }
};