Maximal Square

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 4.

 

int maximalSquare(char** matrix, int matrixRowSize, int matrixColSize) {
    // maintain a maximum be result, and init it to be 0;
    if (matrixRowSize == 0 || matrixColSize == 0)
        return 0;
    // memorize a array
    int **result = (int **)malloc(matrixRowSize * sizeof(int *));
    for (int i = 0; i < matrixRowSize; i++){
        result[i] = (int *)malloc(matrixColSize * sizeof(int));
    }
    int s_max = matrix[0][0] - '0';
    //find the max from first row and first col
    for (int i = 0; i < matrixRowSize; i++){
        result[i][0] = matrix[i][0] - '0';
        if (result[i][0] == 1)
            s_max = 1;
    }
    for (int i = 1; i < matrixColSize; i++){
        result[0][i] = matrix[0][i] - '0';
        if (result[0][i] == 1)
            s_max = 1;
    }
    // go over row form 1 to rowSize
    for (int row = 1; row < matrixRowSize; row++){
        // go over col form 1 to colSize
        for (int col = 1; col < matrixColSize; col++){
            // if this matrix[row][col] be 1
            if (matrix[row][col] == '1' && result[row - 1][col - 1] > 0){
                int check_num = sqrt(result[row - 1][col - 1]);
                // check matrix[row][col - sqrt(i)] to matrix[row][col - 1]
                // and check matrix[row - sqrt(i)] to maxtrix[row - 1][col]
                int i = 0;
                for (; i < check_num; i++){
                    if (result[row][col - 1 - i] == 0 || result[row - 1 - i][col] == 0){
                        break;
                    }
                }
                // if one is 0 ,this one is 1
                // else be (i + 1)^2
                result[row][col] = (i + 1) * (i + 1);
                //max refresh
                if (s_max < result[row][col])
                    s_max = result[row][col];
            }
            else{
                result[row][col] = matrix[row][col] - '0';
            }
        }
    }
    // return max
    return s_max;
}

• 思路是，把每个位置的值保存到另一个数组
• 如果该位置的值是0，则直接复制0
• 如果是1，则判断左上角的值，
• 如果是0，则直接复制
• 否则，判断同行， 同列上的值，是否都为1