[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.

最笨的办法，暴力算，果然超时了！

 1 class Solution {

 2 public:

 3     int largestRectangleArea(vector<int> &height) {

 4         int minHeight;

 5         int maxArea = 0, area;

 6         for (int i = 0; i < height.size(); ++i) {

 7             minHeight = height[i];

 8             for (int j = i; j < height.size(); ++j) {

 9                 minHeight = (minHeight < height[j]) ? minHeight : height[j];

10                 area = minHeight * (j - i + 1);

11                 maxArea = (maxArea > area) ? maxArea : area;

12             }

13         }

14         return maxArea;

15     }

16 };

下面是重点， 可以通过两个栈来保存之前的信息，以减少比较次数，经典就是经典啊。我们要维护两个栈，要注意的是栈中的节点的高度一定是比当前节点小的，若发现height栈顶元素比当前元素大，则要将其出栈，同时计算面积。还有一点要注意的是处理完所有元素后若栈不为空，那么这些元素肯定是从idx延伸到最后的，还要计算其面积。

Actually, we can decrease the complexity by using stack to keep track of the height and start indexes. Compare the current height with previous one.

Case 1: current > previous (top of height stack)
Push current height and index as candidate rectangle start position.

Case 2: current = previous
Ignore.

Case 3: current < previous
Need keep popping out previous heights, and compute the candidate rectangle with height and width (current index - previous index). Push the height and index to stacks.

(Note: it is better use another different example to walk through the steps, and you will understand it better).

 

 1 class Solution {

 2 public:

 3     int largestRectangleArea(vector<int> &height) {

 4         int n = height.size();

 5         stack<int> stkHeight;

 6         stack<int> stkIdx;

 7         int maxArea = 0, tmpArea;

 8         int tmpHeight, tmpIdx;

 9         for (int i = 0; i < n; ++i) {

10             if (stkHeight.empty() || height[i] > stkHeight.top()) {

11                 stkHeight.push(height[i]);

12                 stkIdx.push(i);

13             }

14             else if (height[i] < stkHeight.top()) {

15                 //tmpIdx = 0;

16                 while (!stkHeight.empty() && height[i] <= stkHeight.top()) {

17                     tmpHeight = stkHeight.top();

18                     stkHeight.pop();

19                     tmpIdx = stkIdx.top();

20                     stkIdx.pop();

21                     tmpArea = tmpHeight * (i - tmpIdx);

22                     maxArea = (maxArea > tmpArea) ? maxArea : tmpArea;

23                 }

24                 stkHeight.push(height[i]);

25                 stkIdx.push(tmpIdx);

26             }

27         }

28         while (!stkHeight.empty()) {

29             tmpHeight = stkHeight.top();

30             stkHeight.pop();

31             tmpIdx = stkIdx.top();

32             stkIdx.pop();

33             tmpArea = tmpHeight * (n - tmpIdx);

34             maxArea = (maxArea > tmpArea) ? maxArea : tmpArea;

35         }

36         return maxArea;

37     }

38 };

 上面的代码太臃肿了，可以简化如下：

 1 class Solution {

 2 public:

 3     int largestRectangleArea(vector<int> &line) {

 4         if (line.size() < 1) return 0;

 5         stack<int> S;

 6         line.push_back(0);

 7         int maxArea = 0, tmpArea;

 8         for (int i = 0; i < line.size(); ++i) {

 9             if (S.empty() || line[i] > line[S.top()]) {

10                 S.push(i);

11             } else {

12                 int idx = S.top();

13                 S.pop();

14                 tmpArea = line[idx] * ((S.empty() ? i : i - S.top() - 1));

15                 maxArea = (maxArea > tmpArea) ? maxArea : tmpArea;

16                 --i;

17             }

18         }

19         return maxArea;

20     }

21 };