find the largest subsquare surrounded by ‘w’ 和‘W’

最新推荐文章于 2021-05-27 15:47:46 发布

sjphiChina

最新推荐文章于 2021-05-27 15:47:46 发布

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本文介绍了一种算法，用于在一个包含黑白元素的二维数组中找到最大的全由黑色元素构成的正方形，并返回该正方形的面积。通过动态规划的方法，记录每个位置能够形成的最大黑色正方形边长，最终找出满足条件的最大正方形。

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一个char[][] board，里面有‘b’'B''w''W'四种，bB都表示black，wW都表示white，找最大的正方形面积，这个正方形四个边都是black，正方形中间随便黑白无所谓。Example：一个5x5的board如下，符合条件最大正方形是个3x3的，return面积9。
wwbww
wbbbb
bbWbw
Bbbbb
wwwww

参考：点击打开链接

public static void main(String[] args) {
        int mat1[][] = { { 'b', 'w', 'b', 'b', 'b', 'b' }, { 'b', 'w', 'b', 'b', 'w', 'b' },
                { 'b', 'b', 'b', 'w', 'w', 'b' }, { 'w', 'b', 'b', 'b', 'b', 'b' },
                { 'b', 'b', 'b', 'w', 'b', 'w' }, { 'w', 'w', 'b', 'w', 'w', 'w' } };
        int len = findSubSquare(mat1);
        System.out.println(len);
        int mat2[][] = { { 'w', 'w', 'b', 'w', 'w' }, { 'w', 'b', 'b', 'b', 'b' },
                { 'b', 'b', 'w', 'b', 'w' }, { 'B', 'b', 'b', 'b', 'b' },
                { 'w', 'w', 'w', 'w', 'w', 'w' } };
        len = findSubSquare(mat2);
        System.out.println(len);
    }

    static int getMin(int x, int y) {
        return (x < y) ? x : y;
    }

    // Returns size of maximum size subsquare matrix
    // surrounded by 'X'
    static int findSubSquare(int mat[][]) {
        int max = 1; // Initialize result
        int len = mat.length;
        // Initialize the left-top value in hor[][] and ver[][]
        int[][] hor = new int[len][len];
        int[][] ver = new int[len][len];

        // Fill values in hor[][] and ver[][]
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < len; j++) {
                if (mat[i][j] == 'w' || mat[i][j] == 'W')
                    ver[i][j] = hor[i][j] = 0;
                else {
                    hor[i][j] = (j == 0) ? 1 : hor[i][j - 1] + 1;
                    ver[i][j] = (i == 0) ? 1 : ver[i - 1][j] + 1;
                }
            }
        }

        // Start from the rightmost-bottommost corner element and find
        // the largest ssubsquare with the help of hor[][] and ver[][]
        for (int i = len - 1; i >= 1; i--) {
            for (int j = len - 1; j >= 1; j--) {
                // Find smaller of values in hor[][] and ver[][]
                // A Square can only be made by taking smaller
                // value
                int small = getMin(hor[i][j], ver[i][j]);

                // At this point, we are sure that there is a right
                // vertical line and bottom horizontal line of length
                // at least 'small'.

                // We found a bigger square if following conditions
                // are met:
                // 1)If side of square is greater than max.
                // 2)There is a left vertical line of length >= 'small'
                // 3)There is a top horizontal line of length >= 'small'
                while (small > max) {
                    if (ver[i][j - small + 1] >= small && hor[i - small + 1][j] >= small) {
                        max = small;
                    }
                    small--;
                }
            }
        }
        return max;
    }