zoukankan      html  css  js  c++  java
  • [CareerCup] 18.11 Maximum Subsquare 最大子方形

    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

  • 相关阅读:
    锋利的BFC
    inline和inline-block的间隙问题
    margin和padding的四种写法
    js中Math.round、parseInt、Math.floor和Math.ceil小数取整小结
    使用vscode自动编译less
    redux获取store中的数据
    react显示隐藏动画
    react使用路由
    react中使用fetchjsonp获取数据
    vue兼容到ie9
  • 原文地址:https://www.cnblogs.com/grandyang/p/5484008.html
Copyright © 2011-2022 走看看