106_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.

1:特殊情况；2:设置堆栈，将height中的的索引值保存到stack中，并且索引值指向的数值依次增大；3:在数字出堆栈的时候计算最大值；4:数字完全进入堆栈后，再出堆栈，注意堆栈是否为null的情况

    int largestRectangleArea(vector<int> &height)
    {
        if(height.size() == 0)
        {
            return 0;
        }
        
        int size = (int)height.size();
        stack<int> myStack;
        int maxValue = 0;
        
        for(int i = 0; i < size; i++)
        {
            if(myStack.empty() || height[i] >= height[myStack.top()])
            {
                myStack.push(i);
            }
            else
            {
                while(!myStack.empty() && height[i] < height[myStack.top()])
                {
                    int temp = height[myStack.top()];
                    myStack.pop();
                    
                    int tempValue = temp * (!myStack.empty() ? i - myStack.top() - 1 : i);
                    if(tempValue > maxValue)
                    {
                        maxValue = tempValue;
                    }
                }
                myStack.push(i);
            }
        }
        
        while(!myStack.empty())
        {
            int index = myStack.top();
            int temp = height[index];
            myStack.pop();

            int tempValue = temp * (!myStack.empty() ? size - myStack.top() - 1 : size);
            
            if(tempValue > maxValue)
            {
                maxValue = tempValue;
            }
        }
        return maxValue;
    }