Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
The largest rectangle is shown in the shaded area, which has area = 10 unit.
For example,
Given heights = [2,1,5,6,2,3],
return 10.
package go.jacob.day702;
import java.util.Stack;
public class Demo2 {
/*
* Runtime: 24 ms.Your runtime beats 67.12 % of java submissions.
*/
public int largestRectangleArea(int[] heights) {
if (heights == null || heights.length < 1)
return 0;
Stack<Integer> stack = new Stack<Integer>();
stack.push(-1);
int max = 0;
for (int i = 0; i < heights.length; i++) {
// 把堆栈中比heights[i]大的元素都pop出，并计算面积
while (stack.peek() != -1 && heights[i] < heights[stack.peek()]) {
int top = stack.pop();
// 面积计算公式：(i - 1-stack.peek() ) * heights[top])
max = Math.max(max, (i - 1 - stack.peek()) * heights[top]);
}
stack.push(i);
}
while (stack.peek() != -1) {
int top = stack.pop();
max = Math.max(max, (heights.length - 1 - stack.peek()) * heights[top]);
}
return max;
}
}变形题：maximal rectangle
把每一行作为x轴起点，看成求直方图中最大矩形面积问题
public int maximalRectangle(char[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return 0;
int m = matrix.length, n = matrix[0].length;
int max = 0;
int[] h = new int[n];
for (int i = 0; i < m; i++) {
Stack<Integer> stack = new Stack<Integer>();
stack.push(-1);
for (int j = 0; j < n; j++) {
if (matrix[i][j] == '1')
h[j] += 1;
else
h[j] = 0;
}
for (int j = 0; j < n; j++) {
while (stack.peek() != -1 && h[j] < h[stack.peek()]) {
int top = stack.pop();
max = Math.max(max, (j - 1 - stack.peek()) * h[top]);
}
stack.push(j);
}
while (stack.peek() != -1) {
int top = stack.pop();
max = Math.max(max, (n - 1 - stack.peek()) * h[top]);
}
}
return max;
}