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.

方法一：动态规划，这个思路其实是计算不同长度组合的面积，记录最大的。O(N^2)复杂度，小例子能过，大例子超时。

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = height.size();
        if(n == 0) return 0;
        vector<int> minHeight(n,0);   
        int maxArea = 0;
        
        for(int i = 0;i < n;i++)
        {
            minHeight[i] = height[i];
            if(height[i] > maxArea) maxArea = height[i];
        }
        
        
        for(int l = 2;l <=n ;l++)
        {
            for(int j = n-1;j >= l-1;j--)
            {
                minHeight[j] = min(minHeight[j-1],height[j]);
                int area = minHeight[j]*l;
                if(area > maxArea) maxArea = area;
            }
        }
        

        return maxArea;
    }
};

方法二：

A more straight-forward and easy-to-learn method, no stack:

left[i] is the most-left side column that i can extend to;

    left[i] = max{j: h[j]>h[i] && j<i};

right[i] is the most-right side column that i can reach at;

    right[i] = min{j: h[j]>h[i] && j>i};

so, left[i] can be calculated recursively:

    left[i+1] = left[i],   if h[i+1] == h[i];
              = i,         if h[i+1] > h[i];
              = 0~left[i], if h[i+1] < h[i]; this can be done recursively

such as right[i].

the rectangle area is calculated as: area[i] = (right[i]-left[i]-1) * h[i];

方法三：目前想到的复杂度最小的方法，为O(n)。思路是每次计算以当前height为限制的最大范围的面积，实际和方法二一样。

代码如下：

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

100 milli secs.

看到了更简洁的代码表达：

class Solution {
public:
  int largestRectangleArea(vector<int> &h) {
    stack<int> p;
    int i = 0, m = 0;
    h.push_back(0);
    while(i < h.size()) {
        if(p.empty() || h[p.top()] <= h[i])
            p.push(i++);
        else {
            int t = p.top();
            p.pop();
            m = max(m, h[t] * (p.empty() ? i : i - p.top() - 1 ));
        }
    }
    return m;
}
};

100 milli secs.