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.

题目解析：

方案一：

这道题目想到了动态规划，m[i,j] = min( m[i,j-1]/(j-i) , height[j] ) * (j-i+1)；设置动态设置二维数组，也是以len为主循环。最后遍历整个二维数组，找出最大值。方法正确，但提交的时候，内存超限。还是数组太大造成的。

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int n = height.size();
        int **arr = new int*[n];
        for(int i = 0;i < n;i++)
            arr[i] = new int[n];

        for(int i = 0;i < n;i++)
            arr[i][i] = height[i];
        for(int len = 2;len <= n;len++){
            for(int i = 0;i < n-len+1;i++){
                int j = i + len -1;
                int min = arr[i][j-1]/(j-i) > height[j] ? height[j] : arr[i][j-1]/(j-i);
                arr[i][j] = min * (j-i+1);
            }
        }
        int largest = 0;
        for(int i = 0;i < n;i++)
            for(int j = i;j < n;j++){
                if(largest < arr[i][j]){
                    largest = arr[i][j];
                }
            }

        for(int i = 0;i < n;i++)
            delete[] arr[i];
        delete[] arr;

        return largest;
    }
};

方案二：

链接中用到栈的方法，各种各样，回头有时间了再深究……

http://www.cnblogs.com/remlostime/archive/2012/11/25/2787359.html

http://blog.sina.com.cn/s/blog_727d57100100o3e6.html

http://www.2cto.com/kf/201207/140484.html

http://blog.csdn.net/doc_sgl/article/details/11805519

http://www.cnblogs.com/avril/archive/2013/08/24/3278873.html