[LeetCode] 84. Largest Rectangle in Histogram（直方图里的最大矩形）

• Difficulty: Hard
• Related Topics: Array, Stack
• Link: https://leetcode.com/problems/largest-rectangle-in-histogram/

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.
给定 n 各非负整数，表示直方图中每个柱子的高度（所有柱子的宽度都是 1），找到该立方图中面积最大的矩形的面积。

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
上图中，每个柱子的高度是 1，高度为 [2, 1, 5, 6, 2, 3]。

The largest rectangle is shown in the shaded area, which has area = 10 unit.
上图中面积最大的矩形为阴影所示，面积为 10 个单位。

Example

Input: [2,1,5,6,2,3]
Output: 10

Solution

对于每一个 heights[i]，其最大的矩形面积为以 heights[i] 为高的矩形，也就是说，包含 heights[i] 的矩形范围内不能有比 height[i] 还矮的柱子。于是我们就需要知道每个柱子左右两边第一个比该柱子矮的柱子。常规方法需要 (O(N^2)) 时间找到所有柱子的左右两边第一个比该柱子矮的柱子。但实际上可以利用已经计算的结果。以找 heights[i] 左边第一个比它矮的柱子为例，如果 heights[i - 1] 比 heights[i] 更高，那么第一个比 heights[i] 矮的柱子肯定也是比 heights[i - 1] 矮的柱子。找右边的柱子同理。代码如下：

import kotlin.math.max

class Solution {
    fun largestRectangleArea(heights: IntArray): Int {
        if (heights.isEmpty()) {
            return 0
        }

        val lessFromLeft = IntArray(heights.size)
        val lessFromRight = IntArray(heights.size)
        lessFromLeft[0] = -1
        lessFromRight[lessFromRight.lastIndex] = heights.size

        for (i in 1..heights.lastIndex) {
            var p = i - 1
            while (p >= 0 && heights[p] >= heights[i]) {
                p = lessFromLeft[p]
            }
            lessFromLeft[i] = p
        }

        for (i in (heights.lastIndex - 1) downTo 0) {
            var p = i + 1
            while (p < heights.size && heights[p] >= heights[i]) {
                p = lessFromRight[p]
            }
            lessFromRight[i] = p
        }

        var result = 0
        for (i in heights.indices) {
            result = max(result, heights[i] * (lessFromRight[i] - lessFromLeft[i] - 1))
        }

        return result
    }
}