Leetcode Analysis _84. Largest Rectangle in histogram

"title"

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Given n non-negative integers representing the histogram ' s bar height where the width of each bar are 1, find the area of L Argest rectangle in the histogram.

Above is a histogram where width to each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown into the shaded area and which has area = ten unit.

For example,
Given Heights = [2,1,5,6,2,3],
return "Analysis"

Best Runtime:

Just see this problem when you know you can use dynamic programming, but due to limited capacity, so to the end or did not write an O (n), can only help and other bloggers online, only to find the idea is: Wait until heights[i]<=heights[i-1] The time to calculate, ask Max_height
And there is the heights of a number 0 after the vector, which loops

Class Solution {public
:
    int Largestrectanglearea (vector<int>& heights) {
        heights.push_back (0 );
        int size = Heights.size ();
        stack<int> s;
        int max_area = 0;
        for (int i = 0; i < size i + +) {while
            (!s.empty () && heights[s.top ()] > Heights[i]) {
                int top = S.top ();
                S.pop ();
                Max_area = Max (Max_area, Heights[top] * (I-(S.empty ()? -1:s.top ())-1);
            S.push (i);
        }
        return max_area;
    }
;

Of course, our stacks pop operation can be rewritten using I-compact to avoid always going back and forth pop operations , like this:

Class Solution {public
:
    int Largestrectanglearea (vector<int>& heights) {
        heights.push_back (0 );
        int size = Heights.size ();
        stack<int> s;
        int max_area = 0;
        for (int i = 0; i < size; I + +) {
            if (s.empty () | | heights[s.top ()] <= heights[i]) s.push (i);
            else{
                int top = S.top ();
                S.pop ();
                Max_area = Max (Max_area, Heights[top] * (S.empty () i:i-1-s.top ()));
                I--; 
            }
        }
        return max_area;
    }
;

This time with the if-else structure rather than the while structure, because I-compact, and then in the For loop will also i++ so with the above method of thinking is the same

But the efficiency of the above two solutions has always been between 22ms and 26ms , the runtime on the Leetcode is only 50% , because we always need to push the operation, Each time you need to start a space to push
So we can use vector or array structure, use i++,i– to operate, improve efficiency

Class Solution {public
:
    int Largestrectanglearea (vector<int>& heights) {
        heights.push_back (0 );
        int cur = 0, Max_area = 0, top = 0;
        int * stack = new int[heights.size ()];
        Vector<int> Stack (heights.size (), 0);
        Stack[top] =-1;
        for (int i = 0; i < heights.size (); ++i) {while
            (top>0 && heights[stack[top] >= heights[i]) {
                cur = (i-stack[top-1]-1) *heights[stack[top]];
                top--;
                Max_area = max (cur, max_area);
            }
            Stack[++top] = i;
        }
        return max_area;
    }
;

Efficiency as shown: