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/