leetcode 84. Largest Rectangle in Histogram

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

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

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,
Given heights = [2,1,5,6,2,3],
return 10.

思路，单调栈，维护一个递增的栈。 时间复杂度O(n)

class Solution {
	public:
	struct node {
		int x;
		int left;
		int right;
	};
	int largestRectangleArea(vector<int>& heights) {
		vector<node> st;
		int n = heights.size();
        if (n == 0) return 0;
        if (n == 1) return heights[0];
		vector<int> sum(n);
		for (int i = 0; i < n; ++i) {
			if (i > 0) sum[i] = sum[i-1] + 1; 
            else sum[i] = 1;
		}
        int ans = 0;
		for (int i = 0; i < n; ++i) {
			if (st.empty() || st.back().x < heights[i]) {
				st.push_back({heights[i],i,i});
			} else {
				int ileft = 0, iright = 0;
				while (!st.empty()) {
					node v = st.back();
					if (v.x >= heights[i]) {
						st.pop_back();
						ileft = v.left;
						if (st.size()) st.back().right = v.right;
						int tmp = (sum[v.right]-sum[v.left-1])*v.x;
						if (tmp > ans) ans = tmp;
					} else break;
				}
				st.push_back({heights[i],ileft,i});
			}
			
		}
		while (!st.empty()) {
			node v = st.back();
			st.pop_back();
			if (st.size()) st.back().right = v.right;
			int tmp = (sum[v.right]-sum[v.left-1])*v.x;
			if (tmp > ans) ans = tmp;
		}
		return ans;
	}
};

class Solution {
public:
    
    int largestRectangleArea(vector<int>& heights) {
        heights.push_back(0);
        int n = heights.size();
        stack<int> st;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            while (!st.empty() && heights[i] <= heights[st.top()]) {
                int top = st.top();
                st.pop();
                int l;
                if (st.empty()) l = - 1;
                else l = st.top();
                ans = max(ans, (i - l - 1) * heights[top]); 
            }
            st.push(i);
        }
        return ans;
    }
};