[LeetCode302]Smallest Rectangle Enclosing Black Pixels

Hard ..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.Hide Company Tags GoogleHide Tags Binary Search

如果选中的点是白色，那么最小的面积就是选中的点作为一个边界，加上整个黑色面积。若选中的是黑色，那么最小面积就是整个黑色面积（因为它们连在一起）。

class Solution {public:    int minArea(vector<vector<char>>& image, int x, int y) {        int m = image.size(), n = image[0].size();        int top = searchRow(image, 0, x, 0, n, true);        int bottom = searchRow(image, x+1, m, 0, n, false);        int left = searchCols(image, 0, y, top, bottom, true);        int right = searchCols(image, y+1, n, top, bottom, false);        return (bottom - top) * (right - left);    }    int searchRow(vector<vector<char>>& image, int start, int end, int lo, int hi, bool opt){        while(start != end){            int k = lo, mid = start + (end-start)/2;            while(k < hi && image[mid][k] == '0') ++k;            if(k<hi == opt) end = mid;// k<hi && opt || k>hi && !opt            else start = mid + 1;        }        return start;    }    int searchCols(vector<vector<char>>& image, int start, int end, int lo, int hi, bool opt){        while(start != end){            int k = lo, mid = start + (end - start)/2;            while(k < hi && image[k][mid] == '0') ++k;            if(k < hi == opt) end = mid;            else start = mid + 1;        }        return start;    }};