302. Smallest Rectangle Enclosing Black Pixels

https://leetcode.com/problems/smallest-rectangle-enclosing-black-pixels/description/

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这道题就是2分横轴，再2分纵轴。横轴的时候，传入X进行2分，X那行一定不是全0，然后找到最左边含1的边界。再找最右边含1的边界。
在找左边界时，我们取完中点后如果发现，全0 的话，说明含1边界再中点右侧，反之在左侧。
找右边界时，反一反。
纵轴也是这个思想。

int h;
    int l;
    public int minArea(char[][] recs, int y, int x) {
        h = recs.length;
        l = recs[0].length;
        int[] hor = help(recs,x,true);
        int[] ver = help(recs,y,false);
        return (hor[1]-hor[0]+1)*(ver[1]-ver[0]+1);
    }
    

    private int[] help(char[][] recs, int b,boolean isH) {
        int s = 0, e = b;
        int[] res = new int[2];
        while(s<=e){
            int m = (e-s)/2+s;
            if(allZero(m,recs,isH)){
                s = m+1;
            }else{
                e = m-1;
            }
        }
        res[0] = e+1;
        s = b;
        e = (isH?l:h)-1;
        while(s<=e){
            int m = (e-s)/2+s;
            if(allZero(m,recs,isH)){
                e = m-1;
            }else{
                s = m+1;
            }
        }
        res[1] = s-1;
        return res;
    }
    private boolean allZero(int m, char[][] recs, boolean isH) {
        if(isH){
            for(int i = 0;i < h;i++)
                if(recs[i][m] != '0') return false;
        }
        else{
            for(int j = 0;j < l;j++)
                if(recs[m][j] != '0') return false;
        }
            
        return true;
    }