随想录日记part49

$time：$ 2024.04.20

---

---

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

---

---

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：$O(n)$

---

---

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：$O(n)$