真实的装修公司_源码交易平台网站源码_邯郸seo优化公司_深圳关键词快速排名

---

• 原题链接：84. 柱状图中最大的矩形

---

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。

---

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;}
}

---

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));}
}