leetcode 84 Largest Rectangle in Histogram

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:

1. Forget to record the start_off
2. 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