302. Smallest Rectangle Enclosing Black Pixels-优快云博客

本文介绍了一种算法，用于寻找包含特定黑色像素点的最小矩形区域。该算法通过二分查找确定矩形的边界，适用于黑白二值图像中连通的黑色区域。给出的示例展示了如何使用此算法来解决具体问题。

An image is represented by a binary matrix with 0 as a white pixel and 1 as a black pixel. The black pixels are connected, i.e., there is only one black region. Pixels are connected horizontally and vertically. Given the location (x, y) of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.

For example, given the following image:

[

"0010",

"0110",

"0100"

]

and x = 0, y = 2,

Return 6.

public class Solution {
    public int minArea(char[][] image, int x, int y) {
        if(image == null || image.length == 0) {
            return 0;
        }
        int rowNum = image.length, colNum = image[0].length;
        int left = binarySearch(image, 0, y, 0, rowNum, true, true);
        int right = binarySearch(image, y + 1, colNum, 0, rowNum, true, false);
        int top = binarySearch(image, 0, x, left, right, false, true);
        int bot = binarySearch(image, x + 1, rowNum, left, right, false, false);
        
        return (right - left) * (bot - top);
    }
    
    private int binarySearch(char[][] image, int lo, int hi, int min, int max, boolean searchHorizontal, boolean searchLo) {
        while(lo < hi) {
            int mid = lo + (hi - lo) / 2;
            boolean hasBlackPixel = false;
            for(int i = min; i < max; i++) {
                if((searchHorizontal ? image[i][mid] : image[mid][i]) == '1') {
                    hasBlackPixel = true;
                    break;
                }
            }
            if(hasBlackPixel == searchLo) {
                hi = mid;
            } else {
                lo = mid + 1;
            }
        }
        return lo;
    }
}