代码学习记录49---单调栈

随想录日记part49

$time：$ 2024.04.20

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主要内容：今天开始要学习单调栈的相关知识了，今天的内容主要涉及：柱状图中最大的矩形

• 84.柱状图中最大的矩形

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Topic184.柱状图中最大的矩形

题目：

思路：

代码实现如下：

class Solution {
    public int largestRectangleArea(int[] heights) {
        // 双指针法
        int result = 0;
        int len = heights.length;
        int[] left = new int[len];
        int[] right = new int[len];
        left[0] = -1;
        for (int i = 1; i < len; i++) {
            int t = i - 1;
            while (t >= 0 && heights[t] >= heights[i])
                t = left[t];
            left[i] = t;
        }
        right[len - 1] = len;
        for (int i = len - 2; i >= 0; i--) {
            int t = i + 1;
            while (t < len && heights[t] >= heights[i])
                t = right[t];
            right[i] = t;
        }
        for (int i = 0; i < len; i++) {
            int tem = heights[i] * (right[i] - left[i] - 1);
            result = Math.max(tem, result);
        }
        return result;
    }
}

时间复杂度：$O(n)$
空间复杂度：$O(n)$

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Topic2 接雨水

思路：

与接雨水很像

class Solution {
    public int largestRectangleArea(int[] heights) {
        int result = 0;
        int len = heights.length;
        int[] newheights = new int[len + 2];
        newheights[0] = 0;
        newheights[len + 1] = 0;
        for (int i = 0; i < len; i++) {
            newheights[i + 1] = heights[i];
        }
        heights = newheights;
        Stack<Integer> stack = new Stack<>();
        stack.push(0);
        for (int i = 1; i < len + 2; i++) {
            if (heights[i] > heights[stack.peek()]) {
                stack.push(i);
            } else if (heights[i] == heights[stack.peek()]) {
                stack.pop();
                stack.push(i);
            } else {
                while (!stack.isEmpty() && heights[i] < heights[stack.peek()]) {
                    int mid = stack.pop();
                    if (!stack.isEmpty()) {
                        int h = heights[mid];
                        int w = i - stack.peek() - 1;
                        result = Math.max(h * w, result);
                    }
                }
                stack.push(i);
            }
        }
        return result;
    }
}

class Solution {
    public int trap(int[] height) {
        // 双指针法
        int result = 0;
        int len = height.length;
        for (int i = 0; i < len; i++) {
            if (i == 0 || i == len - 1)
                continue;
            int lheight = height[i];
            int rheight = height[i];
            for (int l = i - 1; l >= 0; l--) {
                lheight = Math.max(lheight, height[l]);
            }
            for (int r = i + 1; r < len; r++) {
                rheight = Math.max(rheight, height[r]);
            }
            int tem = Math.min(rheight, lheight) - height[i];
            if (tem > 0)
                result += tem;
        }
        return result;
    }
}

时间复杂度：$O(n)$
空间复杂度：$O(n)$

原文地址：https://blog.csdn.net/qq_45296815/article/details/138020487

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