leetcode84 Largest Rectangle in Histogram

思路：

使用單調棧計算每一個位置左邊第一個比它矮的位置和右邊第一個比它矮的位置便可。

實現：

 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 }