且构网

分享程序员开发的那些事...
且构网 - 分享程序员编程开发的那些事

[CareerCup] 18.11 Maximum Subsquare 最大子方形

更新时间:2022-09-16 19:25:20

18.11 Imagine you have a square matrix, where each cell (pixel) is either black or white. Design an algorithm to find the maximum subsquare such that all four borders are filled with black pixels.

LeetCode上的原题,请参见我之前的解法Maximal Square。书上给了两种解法,但是比较长:

解法一:

class Subsquare {
public:
    int row, col, size;
    Subsquare(int r, int c, int sz): row(r), col(c), size(sz) {}
    void print() {
        cout << "(" << row << ", " << col << ", " << size << ")" << endl;
    }
};

bool is_square(vector<vector<int>> &matrix, int row, int col, int size) {
    for (int j = 0; j < size; ++j) {
        if (matrix[row][col + j] == 1) return false;
        if (matrix[row + size - 1][col + j] == 1) return false;
    }
    for (int i = 1; i < size - 1; ++i) {
        if (matrix[row + i][col] == 1) return false;
        if (matrix[row + i][col + size - 1] == 1) return false;
    }
    return true;
}

Subsquare* find_square_with_size(vector<vector<int>> &matrix, int squareSize) {
    int cnt = matrix.size() - squareSize + 1;
    for (int row = 0; row < cnt; ++row) {
        for (int col = 0; col < cnt; ++col) {
            if (is_square(matrix, row, col, squareSize)) {
                return new Subsquare(row, col, squareSize);
            }
        }
    }
    return NULL;
}

Subsquare* find_square(vector<vector<int>> &matrix) {
    for (int i = matrix.size(); i >= 1; --i) {
        Subsquare *square = find_square_with_size(matrix, i);
        if (square) return square;
    }
    return NULL;
}

解法二:

class Subsquare {
public:
    int row, col, size;
    Subsquare(int r, int c, int sz): row(r), col(c), size(sz) {}
    void print() {
        cout << "(" << row << ", " << col << ", " << size << ")" << endl;
    }
};

class SquareCell {
public:
    int zerosRight = 0, zerosBelow = 0;
    SquareCell(int right, int below): zerosRight(right), zerosBelow(below){}
    void setZerosRight(int right) {
        zerosRight = right;
    }
    void setZerosBelow(int below) {
        zerosBelow = below;
    }
};

bool is_square(vector<vector<SquareCell*>> &matrix, int row, int col, int size) {
    SquareCell *topLeft = matrix[row][col];
    SquareCell *topRight = matrix[row][col + size - 1];
    SquareCell *bottomRight = matrix[row + size - 1][col];
    if (topLeft->zerosRight < size) return false;
    if (topLeft->zerosBelow < size) return false;
    if (topRight->zerosBelow < size) return false;
    if (bottomRight->zerosRight < size) return false;
    return true;
}

vector<vector<SquareCell*>> process_square(vector<vector<int>> &matrix) {
    vector<vector<SquareCell*>> res(matrix.size(), vector<SquareCell*>(matrix.size()));
    for (int r = matrix.size() - 1; r >= 0; --r) {
        for (int c = matrix.size() - 1; c >= 0; --c) {
            int rightZeros = 0, belowZeros = 0;
            if (matrix[r][c] == 0) {
                ++rightZeros;
                ++belowZeros;
                if (c + 1 < matrix.size()) {
                    SquareCell *pre = res[r][c + 1];
                    rightZeros += pre->zerosRight;
                }
                if (r + 1 < matrix.size()) {
                    SquareCell *pre = res[r + 1][c];
                    belowZeros += pre->zerosBelow;
                }
            }
            res[r][c] = new SquareCell(rightZeros, belowZeros);
        }
    }
    return res;
}

Subsquare* find_square_with_size(vector<vector<SquareCell*>> &processed, int square_size) {
    int cnt = processed.size() - square_size + 1;
    for (int row = 0; row < cnt; ++row) {
        for (int col = 0; col < cnt; ++col) {
            if (is_square(processed, row, col, square_size)) {
                return new Subsquare(row, col, square_size);
            }
        }
    }
    return NULL;
}

Subsquare* find_square(vector<vector<int>> &matrix) {
    vector<vector<SquareCell*>> processed = process_square(matrix);
    // cout << "here" << endl;
    for (int i = matrix.size(); i >= 1; --i) {
        Subsquare *square = find_square_with_size(processed, i);
        if (square) return square;
    }
    return NULL;
}

本文转自博客园Grandyang的博客,原文链接: 最大子方形[CareerCup] 18.11 Maximum Subsquare,如需转载请自行联系原博主。