LeetCode(84) Largest Rectangle in Histogram

题目

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.

LeetCode(84) Largest Rectangle in Histogram_第1张图片
Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

LeetCode(84) Largest Rectangle in Histogram_第2张图片
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.

分析

如木桶理论,题目给定一组矩形柱高度序列,求出其构成的直方图最大面积;

方法一:
依次遍历每个坐标位置,从该柱形左右延展,当高度降低时停止延展,求出其组成的矩形面积,实时更新最大面积;

但是大数据集会超时;

方法二:
查看资料看到一个经典的栈解决方法参考网址~~

给出详细解析:

1、如果已知height数组是升序的,应该怎么做?

比如1,2,5,7,8

那么就是(1*5) vs. (2*4) vs. (5*3) vs. (7*2) vs. (8*1)

也就是max(height[i]*(size-i))

2、使用栈的目的就是构造这样的升序序列,按照以上方法求解。

但是height本身不一定是升序的,应该怎样构建栈?

比如2,1,5,6,2,3

(1)2进栈。s={2}, result = 0

(2)1比2小,不满足升序条件,因此将2弹出,并记录当前结果为2*1=2。

将2替换为1重新进栈。s={1,1}, result = 2

(3)5比1大,满足升序条件,进栈。s={1,1,5},result = 2

(4)6比5大,满足升序条件,进栈。s={1,1,5,6},result = 2

(5)2比6小,不满足升序条件,因此将6弹出,并记录当前结果为6*1=6。s={1,1,5},result = 6

2比5小,不满足升序条件,因此将5弹出,并记录当前结果为5*2=10(因为已经弹出的5,6是升序的)。s={1,1},result = 10

2比1大,将弹出的5,6替换为2重新进栈。s={1,1,2,2,2},result = 10

(6)3比2大,满足升序条件,进栈。s={1,1,2,2,2,3},result = 10

栈构建完成,满足升序条件,因此按照升序处理办法得到上述的max(height[i]*(size-i))=max{3*1, 2*2, 2*3, 2*4, 1*5, 1*6}=8<10

综上所述,result=10

AC代码

class Solution {
public:
    /*方法一:每个坐标点左右延伸(当高度降低停止延伸)构造矩形,但是大集合TLE*/
    int largestRectangleArea1(vector<int> &height) {
        if (height.empty())
            return 0;
        int maxArea = 0;
        int len = height.size();
        for (int i = 0; i < len; ++i)
        {
            /*记录包含第i个柱体的矩形面积*/
            int tmpArea = height[i];
            int left = i - 1, right = i + 1;
            /*左侧扩展*/
            while (left >= 0 && height[left] >= height[i])
            {
                tmpArea += height[i];
                --left;
            }//while
            /*右侧扩展*/
            while (right < len && height[right] >= height[i])
            {
                tmpArea += height[i];
                ++right;
            }//while

            if (maxArea < tmpArea)
                maxArea = tmpArea;
        }//for

        return maxArea;
    }
    /*方法二:利用栈*/
    int largestRectangleArea(vector<int> &height) {
        if (height.empty())
            return 0;
        stack<int> stk;
        int len = height.size();
        int maxArea = 0;
        for (int i = 0; i < len; i++)
        {
            if (stk.empty() || stk.top() <= height[i])
                stk.push(height[i]);
            else
            {
                int count = 0;
                while (!stk.empty() && stk.top() > height[i])
                {
                    count++;
                    maxArea = max(maxArea, stk.top()*count);
                    stk.pop();
                }
                while (count--)
                    stk.push(height[i]);
                stk.push(height[i]);
            }//else
        }//for
        int count = 1;
        while (!stk.empty())
        {
            maxArea = max(maxArea, stk.top()*count);
            stk.pop();
            count++;
        }//while
        return maxArea;
    }
};

GitHub测试程序源码

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