Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing only 1‘s and return its area.
Example:
Input:
[
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]
Output: 6
Iterative的largest rectangle in histogram.
思路：一层一层遍历，到i层时，第i层是1的位置可以向上延伸所有连续的1作为一个直方条条，按这样的规律可以把matrix[0:i][:]看做一个直方图，然后去统计当前情况下的最大rectangle，得到答案后去打擂台。所有层遍历完了，答案就出来了。
直方图的储存：用int[] heights[colLength]来储存，更新的方法是，扫matrix里新的一行时，如果看到’0’就清空heights[j]，如果看到’1’就让heights[j]++。
解释计算直方图里的清空操作：直方条的定义是底部非空向上生长。因为histogram问题里能用stack解决的原因就是，所有直方图最底部开始都是非空的，那么算面积是可以只在意顶上高到哪里，不用担心底部有没有悬空，从而记录高度即可。如果你把matrix的局部转化成直方图的时候看到上面有1但底部是0，那这一列都不合格直方条的定义了。上面的1不用担心，你之前遍历到前面那行的时候算过了。
相关题目：Largest Rectangle in Histogram。 https://www.cnblogs.com/jasminemzy/p/9764297.html
实现：
class Solution {
public int maximalRectangle(char[][] matrix) {
// invalid input.
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return 0;
}
int ans = 0;
int[] heights = new int[matrix[0].length];
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == ‘0‘) {
heights[j] = 0;
} else {
heights[j]++;
}
}
ans = Math.max(ans, maxRecInHistogram(heights));
}
return ans;
}
private int maxRecInHistogram(int[] heights) {
int ans = 0;
Stack<Integer> stack = new Stack<>();
for (int i = 0; i <= heights.length; i++) {
while (!stack.isEmpty() && (i == heights.length || heights[i] < heights[stack.peek()])) {
int height = heights[stack.pop()];
int width = stack.isEmpty() ? i : i - stack.peek() - 1;
ans = Math.max(ans, height * width);
}
stack.push(i);
}
return ans;
}
}
原文地址：https://www.cnblogs.com/jasminemzy/p/9770341.html