【重点】【单调栈】84.柱状图中最大矩形

原创已于 2025-08-10 02:03:49 修改 · 567 阅读

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#单调栈

于 2023-12-25 01:13:04 首次发布

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文章介绍了两种利用单调栈解决最大矩形面积问题的方法，分别是原版和优化版，时间复杂度均为O(N)。重点讲解了如何使用栈来维护每个高度的左右边界，以找到最大的矩形面积。

题目

法1：单调栈[原版]

O(N)+O(N)
低保算法，必须掌握算法！！！

Python

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        res, n = 0, len(heights)
        leftMin = [-1] * n
        rightMin = [n] * n
        stack = [] # 保存索引
        for i in range(n): # 找左侧第1个 < heights[i]的索引
            cur = heights[i]
            while stack and heights[stack[-1]] >= cur:
                stack.pop()
            leftMin[i] = stack[-1] if stack else -1
            stack.append(i)
        
        stack = []
        for i in range(n-1, -1, -1): # 找右侧第1个 < heights[i]的索引
            cur = heights[i]
            while stack and heights[stack[-1]] >= cur:
                stack.pop()
            rightMin[i] = stack[-1] if stack else n
            stack.append(i)
        
        for i in range(n):
            res = max(res, (rightMin[i] - leftMin[i] - 1) * heights[i])
        return res

Java

class Solution {
    public int largestRectangleArea(int[] heights) {
        int n = heights.length, res = 0;
        int[] leftMin = new int[n], rightMin = new int[n];
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < n; ++i) {
            while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {
                stack.pop();
            }
            leftMin[i] = stack.isEmpty() ? -1 : stack.peek();
            stack.push(i);
        }
        stack.clear();
        for (int i = n - 1; i >= 0; --i) {
            while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {
                stack.pop();
            }
            rightMin[i] = stack.isEmpty() ? n : stack.peek();
            stack.push(i);
        }

        for (int i = 0; i < n; ++i) {
            res = Math.max(res, (rightMin[i] - leftMin[i] - 1) * heights[i]);
        }

        return res;
    }
}

法2：单调栈[优化版]

O(N)+O(N)
参考答案
在这里插入图片描述

class Solution {
    public int largestRectangleArea(int[] heights) {
        int n = heights.length, res = 0;
        int[] leftMin = new int[n], rightMin = new int[n];
        Arrays.fill(rightMin, n); // 一定注意这次需要初始化！！！
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < n; ++i) {
            while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {
                rightMin[stack.peek()] = i;
                stack.pop();
            }
            leftMin[i] = stack.isEmpty() ? -1 : stack.peek();
            stack.push(i);
        }

        for (int i = 0; i < n; ++i) {
            res = Math.max(res, (rightMin[i] - leftMin[i] - 1) * heights[i]);
        }

        return res;
    }
}