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‎README.adoc

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|Easy
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//|{counter:codes}
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//|{leetcode_base_url}/largest-rectangle-in-histogram/[84. Largest Rectangle in Histogram^]
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//|{source_base_url}/_0084_LargestRectangleInHistogram.java[Java]
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//|{doc_base_url}/0084-largest-rectangle-in-histogram.adoc[题解]
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//|Hard
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//|
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//
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|{counter:codes}
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|{leetcode_base_url}/largest-rectangle-in-histogram/[84. Largest Rectangle in Histogram^]
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|{source_base_url}/_0084_LargestRectangleInHistogram.java[Java]
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|{doc_base_url}/0084-largest-rectangle-in-histogram.adoc[题解]
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|Hard
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|
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//|{counter:codes}
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//|{leetcode_base_url}/maximal-rectangle/[85. Maximal Rectangle^]
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//|{source_base_url}/_0085_MaximalRectangle.java[Java]

‎docs/0084-largest-rectangle-in-histogram.adoc

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[#0084-largest-rectangle-in-histogram]
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= 84. Largest Rectangle in Histogram
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= 84. 柱状图中最大的矩形
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{leetcode}/problems/largest-rectangle-in-histogram/[LeetCode - Largest Rectangle in Histogram^]
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https://leetcode.cn/problems/largest-rectangle-in-histogram/[LeetCode - 84. 柱状图中最大的矩形 ^]
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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.
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给定 _n_ 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 `1` 。
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*Example 1:*
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求在该柱状图中，能够勾勒出来的矩形的最大面积。
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image::images/0084-00.png[{image_attr}]
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[.small]#Above is a histogram where width of each bar is 1, given height = `[2,1,5,6,2,3]`.#
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image::images/0084-01.png[{image_attr}]
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[.small]#The largest rectangle is shown in the shaded area, which has area = `10` unit.#
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[subs="verbatim,quotes,macros"]
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----
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*Input:* [2,1,5,6,2,3]
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*Output:* 10
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----
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*Example 2:*
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*示例 1:*
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image:images/0084-03.jpg[]
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----
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Input: heights = [2,1,5,6,2,3]
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Output: 10
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Explanation: The above is a histogram where width of each bar is 1.
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The largest rectangle is shown in the red area, which has an area = 10 units.
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----
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....
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输入：heights = [2,1,5,6,2,3]
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输出：10
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解释：最大的矩形为图中红色区域，面积为 10
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....
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*Example 3:*
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*示例 2：*
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image:images/0084-04.jpg[]
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----
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Input: heights = [2,4]
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Output: 4
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----
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....
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输入： heights = [2,4]
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输出： 4
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....
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*提示：*
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* `1 \<= heights.length \<=10^5^`
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* `0 \<= heights[i] \<= 10^4^`
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== 思路分析
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image::images/0084-10.png[{image_attr}]
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单调递增栈
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当需要出栈的时候，就是当前元素小于栈顶元素。这样，弹出栈顶元素，以弹出的栈顶元素作为高，由于是单调递增栈，现在栈顶元素是小于已弹出的栈顶元素。所以，栈顶元素和当前坐标就是以弹出的栈顶元素的左右两个更低的柱子。那么，就弹出的栈顶元素作为高，当前坐标-栈顶元素（存的坐标）-1作为宽，就可以计算相关柱子组成的面积了。
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哨兵技巧非常巧妙。即可减少栈的非空判断，又可以推进剩余元素的计算（如果没有最后的零点哨兵，栈里单调递增的元素最后还要单独处理。）
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[[src-0084]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0084_LargestRectangleInHistogram.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0084_LargestRectangleInHistogram_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/largest-rectangle-in-histogram/solutions/142012/bao-li-jie-fa-zhan-by-liweiwei1419/[84. 柱状图中最大的矩形 - 暴力解法、栈（单调栈、哨兵技巧）^] -- 哨兵技巧牛逼
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. https://leetcode.cn/problems/largest-rectangle-in-histogram/solutions/266844/zhu-zhuang-tu-zhong-zui-da-de-ju-xing-by-leetcode-/[84. 柱状图中最大的矩形 - 官方题解^] -- 看的稀里糊涂。

‎docs/images/0084-10.png

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‎docs/index.adoc

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// include::0084-largest-rectangle-in-histogram.adoc[leveloffset=+1]
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include::0084-largest-rectangle-in-histogram.adoc[leveloffset=+1]
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// include::0085-maximal-rectangle.adoc[leveloffset=+1]
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‎logbook/202503.adoc

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|{doc_base_url}/0074-search-a-2d-matrix.adoc[题解]
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|✅ 二分查找。把矩阵按行“拼接”，然后二分查找，解法非常妙。
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|{counter:codes2503}
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|{leetcode_base_url}/largest-rectangle-in-histogram/[84. 柱状图中最大的矩形^]
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|{doc_base_url}/0084-largest-rectangle-in-histogram.adoc[题解]
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|❌ 单调栈。一脸懵逼。哨兵技巧非常巧妙。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0083_RemoveDuplicatesFromSortedList.java

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/**
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* Runtime: 1 ms, faster than 22.08% of Java online submissions for Remove Duplicates from Sorted List.
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* Memory Usage: 40 MB, less than 7.14% of Java online submissions for Remove Duplicates from Sorted List.
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*
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* @author D瓜哥 · https://www.diguage.com
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* @since 2020-02-04 22:39
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*/
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public ListNode deleteDuplicates(ListNode head) {
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ListNode current = head.next;

‎src/main/java/com/diguage/algo/leetcode/_0084_LargestRectangleInHistogram.java

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package com.diguage.algo.leetcode;
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import java.util.ArrayDeque;
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import java.util.Deque;
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public class _0084_LargestRectangleInHistogram {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-04-23 20:07:47
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*/
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public int largestRectangleArea(int[] heights) {
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int length = heights.length;
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if (length == 1) {
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return heights[0];
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}
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int[] newHeights = new int[length + 2];
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// 两端是两个哨兵
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newHeights[0] = 0;
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newHeights[length + 1] = 0;
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System.arraycopy(heights, 0, newHeights, 1, length);
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heights = newHeights;
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int result = 0;
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length += 2;
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Deque<Integer> stack = new ArrayDeque<>();
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stack.push(0); // 哨兵，可以省去栈的非空判断
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for (int i = 1; i < length; i++) {
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// 单调递增栈
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// 当需要出栈的时候，就是当前元素小于栈顶元素。
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// 这样，弹出栈顶元素，以弹出的栈顶元素作为高，
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// 由于是单调递增栈，现在栈顶元素是小于已弹出的栈顶元素。
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// 所以，栈顶元素和当前坐标就是以弹出的栈顶元素的左右两个更低的柱子
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// 那么，就弹出的栈顶元素作为高，当前坐标-栈顶元素（存的坐标）-1作为宽
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// 就可以计算相关柱子组成的面积了。
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while (heights[i] < heights[stack.peekLast()]) {
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int currHeight = heights[stack.pollLast()];
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int currWidth = i - stack.peekLast() - 1;
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result = Math.max(result, currHeight * currWidth);
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}
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stack.addLast(i);
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}
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return result;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _0084_LargestRectangleInHistogram()
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.largestRectangleArea(new int[]{2, 1, 5, 6, 2, 3});
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}
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}

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