[LeetCode] Largest Rectangle in Histogram

最新推荐文章于 2019-04-10 21:02:20 发布

转载最新推荐文章于 2019-04-10 21:02:20 发布 · 39 阅读

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4776921.html

本文介绍了一种利用栈优化计算直方图中最大矩形面积的方法，相较于使用向量，此方法通常能提高4到8毫秒的运行速度。

This link has a very neat code, which is rewritten below using stack since the push and pop operations of it are O(1) time, while the pop_back and push_back of vector tend to be more time-consuming. This is also verified by the running time of the code on the OJ: stack version is generally 4ms to 8ms faster than vector version.

 1 class Solution {
 2 public:
 3     int largestRectangleArea(vector<int>& height) {
 4         height.push_back(0);
 5         int n = height.size(), area = 0;
 6         stack<int> indexes;
 7         for (int i = 0; i < n; i++) {
 8             while (!indexes.empty() && height[indexes.top()] > height[i]) {
 9                 int h = height[indexes.top()]; indexes.pop();
10                 int l = indexes.empty() ? -1 : indexes.top();
11                 area = max(area, h * (i - l - 1));
12             }
13             indexes.push(i);
14         }
15         return area; 
16     }
17 };

Moreover, it would be better to keep height unmodified. So we loop for n + 1 times and manually set the h = 0 when i == n. The code is as follows.

 1 class Solution {
 2 public:
 3     int largestRectangleArea(vector<int>& height) {
 4         int n = height.size(), area = 0, h, l;
 5         stack<int> indexes;
 6         for (int i = 0; i <= n; i++) {
 7             while (i == n || (!indexes.empty() && height[indexes.top()] > height[i])) {
 8                 if (i == n && indexes.empty()) h = 0, i++;
 9                 else h = height[indexes.top()], indexes.pop();          
10                 l = indexes.empty() ? -1 : indexes.top();
11                 area = max(area, h * (i - l - 1));
12             }
13             indexes.push(i);
14         }
15         return area;
16     }
17 };

转载于:https://www.cnblogs.com/jcliBlogger/p/4776921.html