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_85_maximalRectangle.java
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_85_maximalRectangle.java
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package pp.arithmetic.leetcode;
import java.util.Arrays;
/**
* Created by wangpeng on 2019-07-18.
* 85. 最大矩形
* <p>
* 给定一个仅包含 0 和 1 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。
* <p>
* 示例:
* <p>
* 输入:
* [
* ["1","0","1","0","0"],
* ["1","0","1","1","1"],
* ["1","1","1","1","1"],
* ["1","0","0","1","0"]
* ]
* 输出: 6
* <p>
* 来源:力扣(LeetCode)
* 链接:https://leetcode-cn.com/problems/maximal-rectangle
* 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
*/
public class _85_maximalRectangle {
public static void main(String[] args) {
_85_maximalRectangle maximalRectangle = new _85_maximalRectangle();
int i = maximalRectangle.maximalRectangle(new char[][]{
{'1', '0', '1', '0', '0'},
{'1', '0', '1', '1', '1'},
{'1', '1', '1', '1', '1'},
{'1', '0', '0', '1', '0'}
});
System.out.println(i);
}
/**
* 官方题解:动态规划
*
* @param matrix
* @return
*/
public int maximalRectangle(char[][] matrix) {
if (matrix.length == 0) return 0;
int m = matrix.length;
int n = matrix[0].length;
int[] left = new int[n]; // initialize left as the leftmost boundary possible
int[] right = new int[n];
int[] height = new int[n];
Arrays.fill(right, n); // initialize right as the rightmost boundary possible
int maxarea = 0;
for (int i = 0; i < m; i++) {
int cur_left = 0, cur_right = n;
// update height
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '1') height[j]++;
else height[j] = 0;
}
// update left
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '1') left[j] = Math.max(left[j], cur_left);
else {
left[j] = 0;
cur_left = j + 1;
}
}
// update right
for (int j = n - 1; j >= 0; j--) {
if (matrix[i][j] == '1') right[j] = Math.min(right[j], cur_right);
else {
right[j] = n;
cur_right = j;
}
}
// update area
for (int j = 0; j < n; j++) {
maxarea = Math.max(maxarea, (right[j] - left[j]) * height[j]);
}
}
return maxarea;
}
}