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 shaded area, which has area = 10 unit.

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

　　简单解法：对于每一个height[i],寻找以此height[i]作为长方形的宽，然后寻找此长方形的最左边和最右边。8 milli secs过小数据，大数据超时

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = height.size();
        if(len < 1) return 0;
        int left, right ,i, maxArea = 0;
        for(i = 0; i< len ; i++)
        {
          int  currentHeight =  height[i] ;
          if(currentHeight == 0 ) continue;
          
           left = i;
           while(left >= 0 &&  height[left] >= currentHeight)
           {
                 left --;
           }
           left++;
           right = i;
           
           while(right < len && height[right] >= currentHeight)
           {
                  right++;
           }
           right--;
           
           int area = currentHeight *(right - left + 1) ;
           maxArea = maxArea <  area ? area : maxArea ; 
        }
        
        return maxArea ;
    }
};

      对上一个算法做出改进：用一个栈来优化对每个height[i]左右边界的寻找，从左到右扫描数组，如果当前栈S为空，则把当前的i值放入栈中；或者如果当前节点的高度大于栈顶节点的高度，也把当前的i值压入栈中（注意这里的不断压栈其实在寻找栈顶节点的矩形的右边界）。当当前节点的高度小于于当前栈顶节点的高度时，此节点便是当前栈中所有节点高度值大于该值的长方形的右边界；剩下的任务就是寻找左边界，栈中每个节点的左边界其实是其栈中的上一个元素。

     这样对于数组中每个元素最多进一次栈，出一次栈。所以时间复杂度为O（n）

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
      int len = height.size();
        if(len < 1) return 0;
        stack<int> myS ;
        int area ,i, tp, maxArea = 0;
        for(i = 0; i< len ;)
        {
           int  currentHeight =  height[i] ;
           if(myS.empty() ||   currentHeight  >=  height[myS.top()] )
            { 
                 myS.push(i);
                 i++;
            }else{
                    tp = myS.top();
                    
                    myS.pop();
                    
                    area =  height[tp] *(myS.empty() ? i : i - myS.top() -1 ) ;
                    
                    maxArea = maxArea <  area ? area : maxArea ; 
                    
            
               }
        }
        
        while(!myS.empty())
        {
              tp = myS.top();
            
            myS.pop();
        
            area = height[tp] *( myS.empty() ? i : i - myS.top() -1  ) ;
                    
            maxArea = maxArea <  area ? area : maxArea ;
            
        }
        return maxArea ;
    }
};

[ reference]: http://www.geeksforgeeks.org/largest-rectangle-under-histogram/