首页 > 其他 > 详细

84. Largest Rectangle in Histogram *HARD* -- 求柱状图中的最大矩形面积

时间:2016-08-24 19:16:49      阅读:199      评论:0      收藏:0      [点我收藏+]

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.

技术分享

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

 

技术分享

The largest rectangle is shown in the shaded area, which has area = 10 unit.

 

For example,
Given heights = [2,1,5,6,2,3],
return 10.

class Solution {
public:
    int largestRectangleArea(vector<int>& heights) {
        int n = heights.size();
        if(0 == n)
            return 0;
        int max = 0, area, i, k;
        stack<int> s;
        heights.push_back(0);
        for(i = 0; i <= n; i++)
        {
            if(s.empty() || heights[i] >= heights[s.top()])
            {
                s.push(i);
                continue;
            }
            k = s.top();
            s.pop();
            area = heights[k] * (0 == s.size() ? i : i - s.top() - 1);
            if(area > max)
                max = area;
            i--;
        }
        return max;
    }
};

维护一个栈,每个bar进栈一次,保持栈内元素递增,若新元素小于栈顶元素,则出栈计算面积。

84. Largest Rectangle in Histogram *HARD* -- 求柱状图中的最大矩形面积

原文:http://www.cnblogs.com/argenbarbie/p/5804195.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!