leetcode84 Largest Rectangle in Histogram

思路：
使用单调栈计算每个位置左边第一个比它矮的位置和右边第一个比它矮的位置即可。
实现：
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 class Solution 
 4 {
 5 public:
 6     int largestRectangleArea(vector<int>& heights) 
 7     {
 8         int n = heights.size();
 9         vector<int> l(n, -1);
10         stack<int> s;
11         s.push(0);
12         for (int i = 1; i < n; i++)
13         {
14             if (heights[s.top()] < heights[i])
15             {
16                 l[i] = s.top();
17                 s.push(i);
18             }
19             else
20             {
21                 while (!s.empty() && heights[s.top()] >= heights[i])
22                     s.pop();
23                 if (!s.empty()) l[i] = s.top();
24                 s.push(i);
25             }
26         }
27         while (!s.empty()) s.pop();
28         s.push(n - 1);
29         vector<int> r(n, n);
30         for (int i = n - 2; i >= 0; i--)
31         {
32             if (heights[s.top()] < heights[i])
33             {
34                 r[i] = s.top();
35                 s.push(i);
36             }
37             else
38             {
39                 while (!s.empty() && heights[s.top()] >= heights[i])
40                     s.pop();
41                 if (!s.empty()) r[i] = s.top();
42                 s.push(i);
43             }
44         }
45         int ans = 0;
46         for (int i = 0; i < n; i++)
47         {
48             ans = max(ans, heights[i] * (r[i] - l[i] - 1));
49         }
50         return ans;
51     }
52 };
53 int main()
54 {
55     int a[] = {2, 1, 5, 6, 2, 3};
56     vector<int> v(a, a + 6);
57     cout << Solution().largestRectangleArea(v) << endl;
58     return 0;
59 }