题目来源

力扣2865美丽塔I

题目概述

给你一个长度为 n 下标从 0 开始的整数数组 maxHeights 。

你的任务是在坐标轴上建 n 座塔。第 i 座塔的下标为 i ，高度为 heights[i] 。

如果以下条件满足，我们称这些塔是 美丽 的：

1 <= heights[i] <= maxHeights[i] heights 是一个 山脉 数组。

解题思路

思路一：遍历一遍数组，假设每个位置为顶峰，分别计算该位置为顶峰时的山脉高度，并记录最大值； 思路二：使用两个数组（栈或者双端队列都行，最好是栈，方便编码）一个记录从左到右递增序列在某个位置的最大高度和，另一个记录从右到左递增序列在某个位置的最大高度和，同一个位置的最大高度和相加就是该位置为顶峰的结果。

代码实现

java实现

public class Solution {public long maximumSumOfHeights(List<Integer> maxHeights) {// 长度int length = maxHeights.size();// 构造一座从左到右逐渐增高的山脉数组int[] rightMaxMountain = new int[length];rightMaxMountain[0] = maxHeights.get(0);// 表示记某个位置为顶峰,从左递增山脉的累计高度long[] rightMaxMountainHeightSum = new long[length];rightMaxMountainHeightSum[0] = rightMaxMountain[0];// 表示已构造山脉长度for (int i = 1; i < length; i++) {// 当前高度int current = maxHeights.get(i);long temp = rightMaxMountainHeightSum[i - 1];// 如果当前高度比已经构造的山高度低，已构造高的部分销毁int tempCount = i ;while (tempCount != 0 && rightMaxMountain[tempCount - 1] > current) {temp -= (rightMaxMountain[tempCount - 1] - current);rightMaxMountain[tempCount - 1] = current;tempCount--;}// 记录最大高度rightMaxMountainHeightSum[i] = temp + (rightMaxMountain[i] = current);}// 构造一座从右到左逐渐增高的山脉数组int[] leftMaxMountain = new int[length];leftMaxMountain[length - 1] = maxHeights.get(length - 1);// 表示记某个位置为顶峰,从右递增山脉的累计高度long[] leftMaxMountainHeightSum = new long[length];leftMaxMountainHeightSum[length - 1] = leftMaxMountain[length - 1];// 结果long result = rightMaxMountainHeightSum[length - 1] + leftMaxMountainHeightSum[length - 1] - maxHeights.get(length - 1);for (int i = length - 2; i >= 0; i--) {// 当前高度int current = maxHeights.get(i);long temp = leftMaxMountainHeightSum[i + 1];// 如果已构造山脉比当前高度高，销毁已构造高的部分int tempCount = i;while (tempCount != length - 1 && current < leftMaxMountain[tempCount + 1]){temp -= (leftMaxMountain[tempCount + 1] - current);leftMaxMountain[tempCount + 1] = current;tempCount++;}// 记录最大高度leftMaxMountainHeightSum[i] = temp + (leftMaxMountain[i] = current);result = Math.max(result, leftMaxMountainHeightSum[i] + rightMaxMountainHeightSum[i] - current);}return result;}
}

c++实现

class Solution {
public:long long maximumSumOfHeights(vector<int>& maxHeights) {// 长度int length = maxHeights.size();// 构造一座从左到右逐渐增高的山脉数组int* rightMaxMountain = new int[length];rightMaxMountain[0] = maxHeights[0];// 表示记某个位置为顶峰,从左递增山脉的累计高度long long* rightMaxMountainHeightSum = new long long[length];rightMaxMountainHeightSum[0] = rightMaxMountain[0];// 表示已构造山脉长度for (int i = 1; i < length; i++) {// 当前高度int current = maxHeights[i];long temp = rightMaxMountainHeightSum[i - 1];// 如果当前高度比已经构造的山高度低，已构造高的部分销毁int tempCount = i;while (tempCount != 0 && rightMaxMountain[tempCount - 1] > current) {temp -= (rightMaxMountain[tempCount - 1] - current);rightMaxMountain[tempCount - 1] = current;tempCount--;}// 记录最大高度rightMaxMountainHeightSum[i] = temp + (rightMaxMountain[i] = current);}// 构造一座从右到左逐渐增高的山脉数组int* leftMaxMountain = new int[length];leftMaxMountain[length - 1] = maxHeights[length - 1];// 表示记某个位置为顶峰,从右递增山脉的累计高度long long* leftMaxMountainHeightSum = new long long[length];leftMaxMountainHeightSum[length - 1] = leftMaxMountain[length - 1];// 结果long long result = rightMaxMountainHeightSum[length - 1] + leftMaxMountainHeightSum[length - 1] - maxHeights[length - 1];for (int i = length - 2; i >= 0; i--) {// 当前高度int current = maxHeights[i];long temp = leftMaxMountainHeightSum[i + 1];// 如果已构造山脉比当前高度高，销毁已构造高的部分int tempCount = i;while (tempCount != length - 1 && current < leftMaxMountain[tempCount + 1]) {temp -= (leftMaxMountain[tempCount + 1] - current);leftMaxMountain[tempCount + 1] = current;tempCount++;}// 记录最大高度leftMaxMountainHeightSum[i] = temp + (leftMaxMountain[i] = current);long long newResult = leftMaxMountainHeightSum[i] + rightMaxMountainHeightSum[i] - current;result = result > newResult ? result : newResult;}return result;}
};

算法题记录