【重点!!!】【单调栈】84.柱状图中最大矩形-优快云博客

本文链接：https://blog.youkuaiyun.com/Allenlzcoder/article/details/135189841

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。

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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:
        n = len(heights)
        leftMin, rightMin = [-1] * n, [n] * n
        stack = list()
        for i in range(n): # 找左侧第1个 < heights[i]的索引
            cur = heights[i]
            while len(stack) > 0 and heights[stack[-1]] >= cur:
                stack.pop()
            if stack:
                leftMin[i] = stack[-1]
            stack.append(i)

        stack.clear()
        for i in range(n-1, -1, -1): # 找右侧第1个 < heights[i]的索引
            cur = heights[i]
            while len(stack) > 0 and heights[stack[-1]] >= cur:
                stack.pop()
            if stack:
                rightMin[i] = stack[-1]
            stack.append(i)
        
        res = -1
        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;
    }
}