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  • [LeetCode] Largest Rectangle in Histogram

    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 };
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  • 原文地址:https://www.cnblogs.com/jcliBlogger/p/4776921.html
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