【力扣】84. 柱状图中最大的矩形

给定 n 个非负整数，用来表示柱状图中各个柱子的高度。每个柱子彼此相邻，且宽度为 1 。求在该柱状图中，能够勾勒出来的矩形的最大面积。

示例 1:

输入：heights = [2,1,5,6,2,3]
输出：10
解释：最大的矩形为图中红色区域，面积为 10

示例 2：

输入： heights = [2,4]
输出： 4

提示：
1 <= heights.length <=$10^5$
0 <= heights[i] <= $10^4$

题解

暴力求解

public static int largestRectangleArea(int[] heights) {
    int sum = 0;
    for (int i = 0; i < heights.length; i++) {
        int left = i;
        int right = i;

        //找当前遍历元素之前第一个比它小的元素
        while (left >= 0) {
            if (heights[left] < heights[i]) {
                break;
            }
            left--;
        }

        //找当前遍历元素之后第一个比它小的元素
        while (right < heights.length) {
            if (heights[right] < heights[i]) {
                break;
            }
            right++;
        }

        int w = right - left - 1;
        int h = heights[i];
        sum = Math.max(sum, w * h);
    }
    return sum;
}

双指针

public class Solution {
    public static int largestRectangleArea(int[] heights) {
        int[] minLeftIndex = new int[heights.length];
        int[] minRightIndex = new int[heights.length];


        // 记录左边第一个小于该柱子的下标
        minLeftIndex[0] = -1;
        for (int i = 1; i < heights.length; i++) {
            int t = i - 1;
            // 这里不是用if，而是不断向右寻找的过程
            while (t >= 0 && heights[t] >= heights[i]) {
                t = minLeftIndex[t];
            }
            minLeftIndex[i] = t;
        }


        // 记录每个柱子右边第一个小于该柱子的下标
        minRightIndex[heights.length - 1] = heights.length;
        for (int i = heights.length - 2; i >= 0; i--) {
            int t = i + 1;
            while (t < heights.length && heights[t] >= heights[i]) {
                t = minRightIndex[t];
            }
            minRightIndex[i] = t;
        }

        /*for (int a : minLeftIndex) {
            System.out.println(a);
        }
        System.out.println("______________________________");

        for (int a : minRightIndex) {
            System.out.println(a);
        }*/

        // 求和
        int result = 0;
        for (int i = 0; i < heights.length; i++) {
            int sum = heights[i] * (minRightIndex[i] - minLeftIndex[i] - 1);
            result = Math.max(sum, result);
        }
        return result;
    }

    public static void main(String[] args) {
        int[] heights = {2, 4, 2};
        System.out.println(largestRectangleArea(heights));
    }
}

单调栈

注意：单调栈是递减的

class Solution {
    int largestRectangleArea(int[] heights) {
        Stack<Integer> st = new Stack<Integer>();
        
        // 数组扩容，在头和尾各加入一个元素，防止只递增或者只递减的数组
        int [] newHeights = new int[heights.length + 2];
        newHeights[0] = 0;
        newHeights[newHeights.length - 1] = 0;
        for (int index = 0; index < heights.length; index++){
            newHeights[index + 1] = heights[index];
        }

        heights = newHeights;
        
        st.push(0);
        int result = 0;
        // 第一个元素已经入栈，从下标1开始
        for (int i = 1; i < heights.length; i++) {
            // 注意heights[i] 是和heights[st.top()] 比较 ，st.top()是下标
            if (heights[i] > heights[st.peek()]) {
                st.push(i);
            } else if (heights[i] == heights[st.peek()]) {
                st.pop(); // 这个可以加，可以不加，效果一样，思路不同
                st.push(i);
            } else {
                while (heights[i] < heights[st.peek()]) { // 注意是while
                    int mid = st.peek();
                    st.pop();
                    int left = st.peek();
                    int right = i;
                    int w = right - left - 1;
                    int h = heights[mid];
                    result = Math.max(result, w * h);
                }
                st.push(i);
            }
        }
        return result;
    }
}