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• 原题链接：84. 柱状图中最大的矩形

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1- 思路

题目识别

• 识别1 ：给定一个数组 heights ，求解柱状图的最大面积

单调栈

• 使用 Stack 来实现，遍历数组，存入下标。
• 本题实现的是一个单调递减栈，目标是找一个以当前为基准，左边第一个比这个小的以及右边第一个比该元素小的元素。

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2- 实现

⭐84. 柱状图中最大的矩形——题解思路

class Solution {
    public int largestRectangleArea(int[] heights) {
        int len = heights.length;
        int[] newH = new int[len+2];

        int index = 1;
        for(int i = 0 ; i < len;i++){
            newH[index++] = heights[i];
        }

        heights = newH;
        len = heights.length;
        Stack<Integer> st = new Stack<>();
        int res = 0;
        st.push(0);
        // 单调栈
        for(int i = 0 ; i < len;i++){
            if(heights[i] >= heights[st.peek()]){
                st.push(i);
            }else{
                while(heights[i] < heights[st.peek()]){
                    int mid = st.peek();
                    st.pop();
                    int left = st.peek();
                    int right = i;
                    int h = heights[mid];
                    int width = right-left-1;
                    res = Math.max(res,h*width);
                }
                st.push(i);
            }
        }

        return res;
    }
}

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3- ACM 实现

public class maxArea {


    public static int maxA(int[] heights){
        int len = heights.length;
        int[] newH = new int[len+2];
        int index = 1;
        for(int i = 1 ; i < len;i++){
            newH[index++] = heights[i];
        }

        heights = newH;
        len = heights.length;
        Stack<Integer> st = new Stack<>();
        int res = 0;
        st.push(0);
        // 单调栈
        for(int i = 0 ; i < len;i++){
            if(heights[i] >= heights[st.peek()]){
                st.push(i);
            }else{
                while(heights[i] < heights[st.peek()]){
                    int mid = st.peek();
                    st.pop();
                    int left = st.peek();
                    int right = i;
                    int h = heights[mid];
                    int width = right-left-1;
                    res = Math.max(res,h*width);
                }
                st.push(i);
            }
        }

        return res;
    }


    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String input = sc.nextLine();
        input = input.replace("[","").replace("]","");
        String[] parts = input.split(",");
        int[] nums = new int[parts.length];
        for(int i = 0 ; i < nums.length;i++){
            nums[i] = Integer.parseInt(parts[i]);
        }
        System.out.println("结果是"+maxA(nums));
    }
}