包裹黑色像素点的最小矩形

http://www.lintcode.com/zh-cn/problem/smallest-rectangle-enclosing-black-pixels/

public class Solution {
    /**
     * @param image: a binary matrix with '0' and '1'
     * @param x: the location of one of the black pixels
     * @param y: the location of one of the black pixels
     * @return: an integer
     */
    public int minArea(char[][] image, int x, int y) {
        // write your code here
//        找四个边
//        左
        int left = 0;
        while (true) {
            boolean has1 = false;
            for (int i = 0; i < image.length; i++) {
                if (image[i][left] == '1') {
                    has1 = true;
                    break;
                }
            }
            if (has1) {
                break;
            } else {
                left++;
            }
        }
//        右
        char[] chars = image[0];
        int right = chars.length - 1;
        while (true) {
            boolean has1 = false;
            for (int i = 0; i < image.length; i++) {
                if (image[i][right] == '1') {
                    has1 = true;
                    break;
                }
            }
            if (has1) {
                break;
            } else {
                right--;
            }
        }
        int top = 0;
        while (true) {
            boolean has1 = false;
            for (int i = 0; i < chars.length; i++) {
                if (image[top][i] == '1') {
                    has1 = true;
                    break;
                }
            }
            if (has1) {
                break;
            } else {
                top++;
            }
        }
        int bottom = image.length - 1;
        while (true) {
            boolean has1 = false;
            for (int i = 0; i < chars.length; i++) {
                if (image[bottom][i] == '1') {
                    has1 = true;
                    break;
                }
            }
            if (has1) {
                break;
            } else {
                bottom--;
            }
        }
        return (right - left + 1) * (bottom - top + 1);
    }
}