C++ & Java LeetCode 302 包含全部黑色像素的最小矩形 三种解法：DFS， BFS，二分查找

c++ 代码： 两种解法 ，深度优先搜索（DFS）， 广度优先搜索（BFS）

#include 
#include 
#include 
#include 
using namespace std;

#define INT_MAX 2147483647
#define INT_MIN (-INT_MAX - 1)

class Solution
{
    int x1 = INT_MAX, x2 = INT_MIN;
    int y1 = INT_MAX, y2 = INT_MIN;

public:
    
    int minArea0(vector> &image, int x, int y) // 解法： 深度优先搜索 DFS
    {
        int r = image.size(), c = image[0].size();

        dfs(x, y, r, c, image);

        for (auto ele : loc)
        {
            x1 = min(x1, ele[0]); // 上边界
            x2 = max(x2, ele[0]); // 下边界
            y1 = min(y1, ele[1]); // 左边界
            y2 = max(y2, ele[1]); // 右边界
        }
        return (x2 - x1 + 1) * (y2 - y1 + 1);
    }

    int minArea(vector> &image, int x, int y) // 解法： 广度优先搜索 BFS
    {
        int m = image.size(), n = image[0].size(), i, j, nx, ny, k;
        vector> dir = {{1, 0}, {0, 1}, {0, -1}, {-1, 0}}; // 上下左右 方向数组
        queue> q;
        q.push({x, y});
        image[x][y] = '0'; // 访问过了
        while (!q.empty())
        {
            i = q.front()[0];
            j = q.front()[1];
            q.pop();
            x1 = min(x1, i); // 上边界
            x2 = max(x2, i); // 下边界
            y1 = min(y1, j); // 左边界
            y2 = max(y2, j); // 右边界
            for (k = 0; k < 4; ++k)
            {
                nx = i + dir[k][0];
                ny = j + dir[k][1];
                if (nx >= 0 && nx < m && ny >= 0 && ny < n && image[nx][ny] == '1')
                {
                    q.push({nx, ny});
                    image[nx][ny] = '0'; // 访问过了
                }
            }
        }
        return (x2 - x1 + 1) * (y2 - y1 + 1);
    }

private:
    vector> loc; // image中所有像素值为1的位置集合列表
    int vis[102][102];
    int dir[4][2] = {
        0, 1,
        1, 0,
        -1, 0,
        0, -1};

    bool check(int x, int y, int r, int c)
    {
        if (x < 0 || y < 0 || x >= r || y >= c)
            return false;
        return true;
    }

    void dfs(int x, int y, int r, int c, vector> &img)
    {
        vis[x][y] = 1;
        loc.push_back({x, y});
        for (int i = 0; i < 4; i++)
        {
            int nx = x + dir[i][0], ny = y + dir[i][1];
            if (check(nx, ny, r, c) && !vis[nx][ny] && img[nx][ny] == '1')
                dfs(nx, ny, r, c, img);
        }
    }
};

int main()
{
    int row = 3;
    int column = 4;
    vector> grid(row, vector(column, 0)); // 初始化row*column二维动态数组，初始化值为0
    char A[3][4] = {{'0', '0', '1', '0'},
                    {'0', '1', '1', '0'},
                    {'0', '1', '0', '0'}};

    for (int m = 0; m < grid.size(); m++)
    {
        for (int n = 0; n < grid[m].size(); n++)
        {
            grid[m][n] = A[m][n];
            cout << grid[m][n] << " ";
        }
        cout << "\n";
    }

    int x = 0, y = 2;
    Solution sol = *new Solution();
    int res = sol.minArea(grid, x, y);
    printf("minArea: %d", res);
    return 0;
}

// 编译：g++ demo.cc -o demo
// 执行： ./demo

java代码：二分查找  确定边界

public class Demo {
    public int minArea(char[][] image, int x, int y) { // 二分查找
        int minRow = searchFirst(image, 0, y, true);
        int maxRow = searchLast(image, y, image[0].length - 1, true);
        int minCol = searchFirst(image, 0, x, false);
        int maxCol = searchLast(image, x, image.length - 1, false);

        return (maxRow - minRow + 1) * (maxCol - minCol + 1);
    }

    private int searchFirst(char[][] image, int start, int end, boolean isVertical) {
        int l = start, r = end;
        while (l + 1 < r) {
            int mid = l + (r - l) / 2;
            if (hasBlack(image, mid, isVertical)) {
                r = mid;
            } else {
                l = mid;
            }
        }

        if (hasBlack(image, l, isVertical)) {
            return l;
        }
        return r;
    }

    private int searchLast(char[][] image, int start, int end, boolean isVertical) {
        int l = start, r = end;
        while (l + 1 < r) {
            int mid = l + (r - l) / 2;
            if (hasBlack(image, mid, isVertical)) {
                l = mid;
            } else {
                r = mid;
            }
        }
        if (hasBlack(image, r, isVertical)) {
            return r;
        }
        return l;
    }

    private boolean hasBlack(char[][] image, int rowOrCol, boolean isVertical) {
        if (isVertical) {
            for (int i = 0; i < image.length; ++i) {
                if (image[i][rowOrCol] == '1') {
                    return true;
                }
            }
        } else {
            for (int i = 0; i < image[0].length; ++i) {
                if (image[rowOrCol][i] == '1') {
                    return true;
                }
            }
        }
        return false;
    }
    
	public static void main(String[] args) {
		
		char[][] map = new char[][] {{'0','0','1','0'},
									 {'0','1','1','0'},
									 {'0','0','1','0'}
								 };
		int x =0, y =2;
		Demo d = new Demo();
		int area = d.minArea(map, x, y);
		System.out.println(area);
	}
}

参考链接：

尝试用二分来解决图像处理问题