leetcode 84 Largest Rectangle in Histogram-优快云博客

本文链接：https://blog.youkuaiyun.com/taoqick/article/details/12879155

本文介绍了一种求解直方图中最大矩形面积的有效算法。通过使用栈来跟踪元素，该算法能在O(n)的时间复杂度内找到给定直方图中最大的矩形区域。文章还提供了C++及Python实现代码示例。

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Given an array of integers heights representing the histogram's bar height where the width of each bar is 1, return the area of the largest rectangle in the histogram.

Example 1:

Input: heights = [2,1,5,6,2,3]
Output: 10
Explanation: The above is a histogram where width of each bar is 1.
The largest rectangle is shown in the red area, which has an area = 10 units.

Example 2:

Input: heights = [2,4]
Output: 4

Constraints:

1 <= heights.length <= 105
0 <= heights[i] <= 104

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This is an excellent problem. Two difficulties in this problem:

Forget to record the start_off
Confuse with problem Leetcode 42 407 Trapping Rain Water I II 双指针

class Solution {
 public:
  int largestRectangleArea(vector<int> &height) {
    // Note: The Solution object is instantiated only once and is reused by each test case.
    stack<int> S;
    int i, sum = 0, cursum = 0;
    height.push_back(0);
    for (i = 0; i < height.size(); ++i) {
      if (S.empty() || (!S.empty() && height[i] >= height[S.top()]))
        S.push(i);    
      else {
        int stackmax = S.top();
        S.pop();
        if (S.empty())
          cursum = height[stackmax] * i;
        else
          cursum = height[stackmax] * (i - S.top() - 1);
        
        if (cursum > sum)
          sum = cursum;
        --i;
      }
    }
    return sum;
  }
};

Python version:

class Solution:
    def largestRectangleArea(self, heights: List[int]) -> int:
        l = len(heights)
        max_area = 0
        sta = []
        for i in range(l):
            start_off = i
            while sta and heights[i] < sta[-1][0]:
                prev_height, start_off = sta.pop()
                max_area = max(max_area, prev_height*(i-start_off))
            sta.append((heights[i],start_off))
        while sta:
            prev_height, start_off = sta.pop()
            max_area = max(max_area, prev_height*(l-start_off))
        return max_area