Largest Rectangle in Histogram

本文介绍了一种高效算法来寻找直方图中最大的矩形面积,通过使用栈来辅助计算,确保了时间复杂度为O(n)。示例输入为高度数组[2,1,5,6,2,3],返回结果为10。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

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.

Java代码:

public class Solution {
    public int largestRectangleArea(int[] height) {
        height = Arrays.copyOf(height, height.length + 1);

        int maxRect = 0;
        Stack<Integer> stack = new Stack<Integer>();
        for(int i = 0; i < height.length; ++i) {
            while (!stack.isEmpty() && height[i] < height[stack.peek()]) {
                int rect = height[stack.pop()] * (stack.isEmpty() ? i : (i-stack.peek()-1));
                maxRect = Math.max(maxRect, rect);
            }
            stack.push(i);
        }

        return maxRect;
    }
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值