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.

思路：

一个比较直观的思路是对于每个直方，分别向左右找到比它小的直方，然后求得从当前位置扩展可得到的最大值。然而这需要O（n^2）的复杂度，其中有很多重复的判断。

可以使用一个栈来实现上述思路。当栈为空或者栈顶元素小于当前值时，把当前值压入栈中，否则的话相当于找到了右边界，再从栈中弹出元素，因为栈内元素是递增的，所以弹出来的相当于是左边界。逐个计算面积，更新当前的max值，直到栈顶元素大于当前元素为止。弹栈时需要注意栈为空的情况。

代码：

    int max(int a, int b){
        if(a > b)
            return a;
        return b;
    }
    int largestRectangleArea(vector<int> &height) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        stack<int> Stack;
        int size = height.size();
        if(size == 0)
            return 0;
        int i = 0;
        int maxRec = 0;
        while(i < size){
            if(Stack.empty() || height[Stack.top()] <= height[i]){
                Stack.push(i++);
            }
            else{
                int t = Stack.top();
                Stack.pop();
                int tmpRec = height[t]*(Stack.empty()?i:i-Stack.top()-1);
                if(tmpRec > maxRec)
                    maxRec = tmpRec;
            }
        }
        while(!Stack.empty()){
            int t = Stack.top();
            Stack.pop();
            int tmpRec = height[t]*(Stack.empty()?i:i-Stack.top()-1);
            if(tmpRec > maxRec)
                maxRec = tmpRec;
        }
        return maxRec;
    }