[LeetCode] 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 shad

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

O（n）的方法：http://www.geeksforgeeks.org/largest-rectangle-under-histogram/

思路：借助栈的push和pop，依次把height里极高点的maxS求出来。

     如果没有遇到极高点，就把height的序号push进栈；

     如果遇到极高点，把极高点的maxS求出来，把极高点pop出栈；

     依次类推！ 具体程序见下面代码：

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
       int maxS = 0;
       stack<int> s;
       int currS, current,low;
       int len = height.size();
       for(int i=0;i<len;)
       {
          if(s.empty() || height[i]>=height[s.top()])
              s.push(i++);

          else
          {
            current = s.top();
            s.pop();
            low = s.empty()?-1:s.top();
            currS = height[current]*(i-low-1);
            if(currS>maxS)
                maxS = currS;
          }//end while 
       }//end for
       while(!s.empty())//此时s中剩下的是逐渐上升的矩形
       {
           current = s.top();
           s.pop();
           low = s.empty()?-1:s.top();
           currS = height[current]*(len - low - 1);
           if(currS>maxS)
                maxS = currS;
       }
       return maxS;
    }
};

下面的代码，和上面的基本思想一致，但是复杂度特别高TimeOut，感受一下：

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int len = height.size();
        int maxS = 0;
        for(int i=0;i<len;i++)
        {
            int newS = CentS(i,height);
            if(maxS<newS)
               maxS = newS;
        
        }
        return maxS;
    }
private:
int CentS(int i,vector<int> &height)
{
   int maxS = height[i];
   int small=i-1,big = i+1;
   while(small>=0 && height[small]>=height[i])
   {
     maxS += height[i];
     small--;
   
   }
   while(big<height.size() && height[big]>=height[i])
   {
    maxS += height[i];
    big++;
   }

  return maxS;
}
};

ed area, which has area = 10 unit.