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Sunday, December 23, 2012

[LeetCode] Largest Rectangle in Histogram 解题报告


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.
» Solve this problem

[解题思路]
对于每一个height,遍历前面所有的height,求取面积最大值。时间复杂度是O(n*n)

[Code]
1:  int largestRectangleArea(vector<int> &height) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      //int result[height.size()];  
5:      int maxV = 0;  
6:      for(int i =0; i< height.size(); i++)  
7:      {  
8:        int minV = height[i];  
9:        for(int j =i; j>=0; j--)  
10:        {  
11:          minV = std::min(minV, height[j]);  
12:          int area = minV*(i-j+1);  
13:          if(area > maxV)  
14:            maxV = area;  
15:        }  
16:      }  
17:      return maxV;  
18:    }  

可以过小数据,但是大数据超时。可以理解,这个解法包含了很多重复计算。一个简单的改进,是只对合适的右边界(峰顶),往左遍历面积。

1:  int largestRectangleArea(vector<int> &height) {   
2:       int maxV = 0;   
3:       for(int i =0; i< height.size(); i++)  
4:       {  
5:            if(i+1 < height.size()   
6:                      && height[i] <= height[i+1]) // if not peak node, skip it  
7:                 continue;  
8:            int minV = height[i];   
9:            for(int j =i; j>=0; j--)   
10:            {   
11:                 minV = std::min(minV, height[j]);   
12:                 int area = minV*(i-j+1);   
13:                 if(area > maxV)   
14:                 maxV = area;   
15:            }   
16:       }  
17:       return maxV;   
18:  }   

这样的话,就可以通过大数据。但是这个优化只是比较有效的剪枝,算法仍然是O(n*n).

想了半天,也想不出来O(n)的解法,于是上网google了一下。
如下图所示,从左到右处理直方,i=4时,小于当前栈顶(及直方3),于是在统计完区间[2,3]的最大值以后,消除掉阴影部分,然后把红线部分作为一个大直方插入。因为,无论后面还是前面的直方,都不可能得到比目前栈顶元素更高的高度了。


这就意味着,可以维护一个递增的栈,每次比较栈顶与当前元素。如果当前元素小于栈顶元素,则入站,否则合并现有栈,直至栈顶元素小于当前元素。结尾入站元素0,重复合并一次。

思路很巧妙。代码实现如下, 大数据 76ms过。
1:     int largestRectangleArea(vector<int> &height) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      int stack[height.size()+1], width[height.size()+1];  
5:      if(height.size() == 0) return 0;  
6:      int top = 0, area = INT_MIN;  
7:      stack[0] = 0;  
8:      width[0] = 0;  
9:      int newHeight;  
10:     for(int i =0; i<= height.size(); i++)  
11:     {  
12:        if(i < height.size()) newHeight = height[i];  
13:        else newHeight = -1;  
14:        if(newHeight>= stack[top])  
15:        {  
16:            stack[++top] = newHeight;  
17:            width[top] = 1;  
18:        }  
19:        else  
20:        {  
21:            int minV = INT_MAX;  
22:            int wid= 0;  
23:            while(stack[top] > newHeight)  
24:            {  
25:               minV = min(minV, stack[top]);  
26:               wid += width[top];  
27:               area = max(area, minV*(wid));  
28:               top--;  
29:             }  
30:             stack[++top] = newHeight;  
31:             width[top] = wid+1;  
32:         }  
33:      }  
34:      return area;       
35:    }  

[总结]
这道题蛮有意思。引入stack的做法相当巧妙。

Update: Refactor code 5/7/2013
评论中Zhongwen Ying的code写的比我post的code简洁多了。把他的code format一下集成进来。

1:  int largestRectangleArea(vector<int> &h) {  
2:       stack<int> S;  
3:       h.push_back(0);  
4:       int sum = 0;  
5:       for (int i = 0; i < h.size(); i++) {  
6:            if (S.empty() || h[i] > h[S.top()]) S.push(i);  
7:            else {  
8:                 int tmp = S.top();  
9:                 S.pop();  
10:                 sum = max(sum, h[tmp]*(S.empty()? i : i-S.top()-1));  
11:                 i--;  
12:            }  
13:       }  
14:       return sum;  
15:  }  


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