Smallest rectangle Enclose Black Pixels

最新推荐文章于 2020-09-25 14:09:59 发布

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最新推荐文章于 2020-09-25 14:09:59 发布

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原文链接：http://www.cnblogs.com/shuashuashua/p/5645609.html

本文介绍了一种利用二分搜索算法寻找图像中指定元素最小包围区域的方法。通过定义边界条件，采用递归方式确定目标区域的左、右、上、下边界，实现了对特定像素点周围非零值矩阵的有效搜索。

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 1 public class Solution {
 2     private char[][] image;
 3     public int minArea(char[][] image, int x, int y) {
 4         if (image.length == 0 || image[0].length == 0) {
 5             return 0;
 6         }
 7         this.image = image;
 8         int n = image.length, m = image[0].length;
 9         int left = searchRow(0, x, 0, m, true);
10         int right = searchRow(x + 1, n, 0, m, false);
11         int top = searchCols(0, y, 0, n, true);
12         int bottom = searchCols(y + 1, m, 0, n, false);
13         return (right - left) * (bottom - top);
14     }
15     
16     private int searchCols(int i, int j, int top, int bottom, boolean bound) {
17         while (i != j) {
18             int k = top;
19             int mid = (i + j) / 2;
20             while (k < bottom && image[k][mid] == '0') {
21                 k++;
22             }
23             if(k < bottom == bound) {
24                 j = mid;
25             } else {
26                 i = mid + 1;
27             }
28         }
29         return i;
30     }   
31     
32     private int searchRow(int i, int j, int left, int right, boolean bound) {
33         while (i != j) {
34             int k = left;
35             int mid = (i + j) / 2;
36             while (k < right && image[mid][k] == '0') {
37                 k++;
38             }
39             if(k < right == bound) {
40                 j = mid;
41             } else {
42                 i = mid + 1;
43             }
44         }
45         return i;
46     }
47 }

Binary search:

1. The bound is defined to search "0" or "1". When search 0 - x, we want to get the most left "1". So k < right mean there is a one, j = mid. But when we do x - n search, we want to get most left "0". So when k < right, we still need to keep searching mid + 1.

转载于:https://www.cnblogs.com/shuashuashua/p/5645609.html