Largest Rectangle in Histogram

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 height = [2,1,5,6,2,3],
return 10.

Analysis

分析：

  • 对于每一个bar 'x', calculate the area with 'x' as the smallest bar
  • 比较所有面积取最大 方法：
  • find index of first smaller than 'x' bar on the left: left index
  • find index of first smaller than 'x' bar on the left: right index 实现：
  • traverse all bars from left to right, maintain a stack of bars
  • every bar pushed to stack once
  • bar popped from stack when a bar of smaller height is seen
  • when bar popped, calculate the area with popped bar as smallest bar
  • current index: right index
  • index of previous item: left index

Code

public class Solution {
    public int largestRectangleArea(int[] height) {
        int len = height.length;
        Stack<Integer> s = new Stack<Integer>();
        int maxArea = 0;
        for(int i = 0; i <= len; i++){
            int h = (i == len ? 0 : height[i]);
            if(s.isEmpty() || h >= height[s.peek()]){
                s.push(i);
            //一旦找到一个height更小的，说明stack top的右边界找到了.此时如果stack empty，那么左边界是index=-1, 如果不是empty，左边界是上一个push在stack里面的index
            //因为在左边比它大bar已经处理过了，剩下的一定是比它小的bar，最上面一个就是离它最近比它小的bar
            }else{
                int tp = s.pop();
                maxArea = Math.max(maxArea, height[tp] * (s.isEmpty() ? i : i - 1 - s.peek()));
                i--;
            }
        }
        return maxArea;
    }
}

Note

  • Not fully understand, need re-think

Reference