时间轴

2026-03-18

init

题目：

P84 柱状图中的最大矩形

https://leetcode.cn/problems/largest-rectangle-in-histogram/description/?envType=study-plan-v2&envId=top-100-liked

这个案例是经典单调栈的场景，仔细想这道题，无非也是找到最左边第一个低于自己的矩形，和最右边第一个低于自己的矩形，其实这是经典The Next Greater问题

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#include <vector>
#include <stack>
using std::vector;
using std::stack;

class Solution {
    public:
        int largestRectangleArea(vector<int> &heights)
        {
                int i, n = heights.size(), max_area = 0;
                vector<int> left(n, 0);
                vector<int> right(n, n - 1);
                stack<int> stk1, stk2;

                for (i = 0; i < n; i++) {
                        while (!stk1.empty() && heights[i] < heights[stk1.top()]) {
                                right[stk1.top()] = i - 1; // 右边界
                                stk1.pop();
                        }
                        stk1.push(i);
                }

                // while (!stk.empty()) {
                //         right[stk.top()] = n - 1;
                // }

                for (i = n - 1; i >= 0; i--) {
                        while (!stk2.empty() && heights[i] < heights[stk2.top()]) {
                                left[stk2.top()] = i + 1; // 左边界
                                stk2.pop();
                        }
                        stk2.push(i);
                }

                // while (!stk.empty()) {
                //         left[stk.top()] = 0;
                // }

                for (i = 0; i < n; i++)
                        max_area = std::max(max_area, (right[i] - left[i] + 1) * heights[i]);

                return max_area;
        }
};

int main()
{
        vector<int> heights = { 2, 1, 5, 6, 2, 3 };
        Solution S;
        S.largestRectangleArea(heights);
}