lintcode(122)直方图最大矩形覆盖

Description:

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.

Explanation:

给出 height = [2,1,5,6,2,3]，返回 10

Solution:

计算以当前值为高的矩形面积，所以要找到它的左右边界。建立栈，存储当前元素在数组中的位置。将数组元素一次推入栈中，如果它比栈顶元素小，则栈顶元素的右边界出现了，现在确定他的左边界。此时将栈顶元素出栈。栈中都是比当前元素小的值，因为比当前值大的都因当前值而出栈，所以，可以借此确定当前矩形的左边界，如果栈为空，说明当前元素是栈中最小值，左边界就是0。知道左右边界之后就可以计算面积了，高度是当前值，宽度是 i 或者 i-stack.peek()-1,为什么再-1？因为是左边界，已经不在矩形范围内了，当前值已经出栈了。

public class Solution {    /**     * @param height: A list of integer     * @return: The area of largest rectangle in the histogram     */    public int largestRectangleArea(int[] height) {        // write your code here        if(height.length == 0) return 0;                int[] high = Arrays.copyOf(height , height.length + 1);                int result = 0;        Stack<Integer> rank = new Stack<Integer>();        for(int i = 0;i<high.length;i++){            while(!rank.isEmpty() && high[i] < high[rank.peek()] ){                int area = high[rank.pop()]*(rank.isEmpty()?i:i-rank.peek() - 1);                result = Math.max(result , area);            }            rank.push(i);        }                return result;    }}