码迷,mamicode.com
首页 > 其他好文 > 详细

[CareerCup] 18.11 Maximum Subsquare 最大子方形

时间:2016-05-12 01:28:41      阅读:177      评论:0      收藏:0      [点我收藏+]

标签:

 

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;
}

 

CareerCup All in One 题目汇总

[CareerCup] 18.11 Maximum Subsquare 最大子方形

标签:

原文地址:http://www.cnblogs.com/grandyang/p/5484008.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!