84. 柱状图中最大的矩形

要求：构造矩形
思路：暴力是在每个矩形处找以它为高度的最大面积，即找它左右第一个比它矮的之间的。单调栈：遇到比栈顶矮的就能计算栈顶了，高的就入栈

class Solution {
public:
    int largestRectangleArea(vector<int> &heights) {
        unsigned long size = heights.size();
        if (size == 1) {
            return heights[0];
        }
        int res = 0;
        stack<int> stk;
        for (int i = 0; i < size; ++i) {
            while (!stk.empty() && heights[stk.top()] > heights[i]) {
                int length = heights[stk.top()];
                stk.pop();
                int weight = i;
                if (!stk.empty()) {
                    weight = i - stk.top() - 1;
                }
                res = max(res, length * weight);
            }
            stk.push(i);
        }
        while (!stk.empty()) {
            int length = heights[stk.top()];
            stk.pop();
            int weight = size;//尾部加0
            if (!stk.empty()) {
                weight = size - stk.top() - 1;//连续两个相等的在这里计算
            }
            res = max(res, length * weight);
        }
        return res;
    }
};