题目来源：

庞果网

题目简介：

给定直方图，每一小块的height由N个非负整数所确定，每一小块的width都为1，请找出直方图中面积最大的矩形。    如下图所示，直方图中每一块的宽度都是1，每一块给定的高度分别是[2,1,5,6,2,3]：    那么上述直方图中，面积最大的矩形便是下图所示的阴影部分的面积，面积= 10单位。    请完成函数largestRectangleArea，实现寻找直方图中面积最大的矩形的功能，如当给定直方图各小块的高度= [2,1,5,6,2,3] ，返回10。

代码（本代码在anycodex在线编程网站上测试通过，Try yourself?）：

#include <stdio.h>

int largestRectangleArea(const int *height,int n) {
    int i = 0;
    int j = 0;
    int count = 1;
    int maxResult = 0;
    int temp;
    for (; i < n; i++)
    {
        count = 1;
        if ( i == 0)       // only check the right one if it is the first element
        {
            if ( *(height + i + 1) >= *(height + i))     //if the right element is bigger or equal than the one
            {
                for ( j = i + 1; j < n; j++)      //then calculate from the right
                {
                    if (*(height + j) >= *(height + i))
                        count++;
                    else
                        break;
                }
            }
            temp = count * *(height + i);
            maxResult = temp > maxResult ? temp: maxResult;
            continue;
        }
        else if ( i == n -1 )                    //only check the left side if it is the last element
        {
            if ( *(height + i - 1) >= *(height + i))     //if the left element is bigger or equal than the one
            {
                for ( j = i - 1; j >= 0; j--)      //then calculate from the left
                {
                    if (*(height + j) >= *(height + i))
                        count++;
                    else
                        break;
                }
            }
            temp = count * *(height + i);
            maxResult = temp > maxResult ? temp: maxResult;
            break;
        }
        else
        {
            if (*(height + i + 1) >= *(height + i) && *(height + i -1) < *(height + i))     //a[i+1] >= a[i] but a[i-1] < a[i]
            {
                for ( j = i + 1; j < n; j++)      //then calculate from the right
                {
                    if (*(height + j) >= *(height + i))
                        count++;
                    else
                        break;
                }
                temp = count * *(height + i);
                maxResult = temp > maxResult ? temp: maxResult;
            }
            else if (*(height + i + 1) < *(height + i) && *(height + i -1) >= *(height + i))     //a[i+1] < a[i] but a[i-1] >= a[i]
            {
                for ( j = i - 1; j >= 0; j--)      //then calculate from the left
                {
                    if (*(height + j) >= *(height + i))
                        count++;
                    else
                        break;
                }
            }
            else if (*(height + i + 1) >= *(height + i) && *(height + i -1) >= *(height + i))     //a[i+1] >= a[i] but a[i-1] >= a[i]
            {
                //printf("a[i] a[i+1] a[i-1] %d %d %d\n", *(p + i), *(p + i + 1), *(p + i -1));
                for ( j = i - 1; j >= 0; j--)      //firstly, calculate from the left
                {
                    if (*(height + j) >= *(height + i))
                        count++;
                    else
                        break;
                }
                for ( j = i + 1; j < n; j++)      //then calculate from the right
                {
                    if (*(height + j) >= *(height + i))
                        count++;
                    else
                        break;
                }
            }
            temp = count * *(height + i);
            maxResult = temp > maxResult ? temp: maxResult;
        }
        //printf("The maxResult in each round is %d\n", maxResult);
    }
    return maxResult;  
}



//start 提示：自动阅卷起始唯一标识，请勿删除或增加。
int main()
{    
    //
    int a[] = {2,1,5,6,3,1};
    int len = sizeof(a)/sizeof(int);
    int result;
    result = largestRectangleArea((const int *)a,len);
    printf("The largest rectangle area is %d\n", result);
    return 0;
}

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