[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:  }