气温变化趋势
难度:
标签:
题目描述
English description is not available for the problem. Please switch to Chinese.
代码结果
运行时间: 30 ms, 内存: 16.3 MB
/*
题目思路:
1. 使用Java Stream处理数组,首先将气温数组转换为相邻两天的气温变化趋势。
2. 比较两个趋势数组,计算相同趋势的连续天数。
3. 使用流的特性来找到最长的连续相同趋势的天数。
*/
import java.util.stream.IntStream;
public class Solution {
public int maxEqualTrendDays(int[] temperatureA, int[] temperatureB) {
int[] trendA = IntStream.range(1, temperatureA.length)
.map(i -> Integer.compare(temperatureA[i], temperatureA[i - 1]))
.toArray();
int[] trendB = IntStream.range(1, temperatureB.length)
.map(i -> Integer.compare(temperatureB[i], temperatureB[i - 1]))
.toArray();
int maxLength = 0, currentLength = 0;
for (int i = 0; i < trendA.length; i++) {
if (trendA[i] == trendB[i]) {
currentLength++;
maxLength = Math.max(maxLength, currentLength);
} else {
currentLength = 0;
}
}
return maxLength;
}
}
解释
方法:
The provided solution aims to find the longest contiguous subarray where the temperature trends between two locations are the same. Firstly, it calculates the daily temperature change for each location and represents it as -1, 0, or 1 (decrease, no change, increase). These trends are stored in two lists (t1 for location A, t2 for location B). Then, by iterating over these trend lists, the solution counts the length of contiguous subarrays where the trends match. The maximum length found during this process is the output.
时间复杂度:
O(n)
空间复杂度:
O(n)
代码细节讲解
🦆
为什么在计算温度趋势时,选择使用-1、0、1来表示温度的变化,而不是直接使用温度差?
▷🦆
在解决方案中,当温度没有变化时(即温度差为0),趋势值设置为0。这样的处理方式是否可能影响结果的准确性,比如在连续多天温度不变的情况下?
▷🦆
代码中通过一个循环来比较两个地点的温度趋势,为什么在发现趋势不匹配时,要将天数计数器`days`重置为0?是否有其他方法可以避免频繁的重置操作?
▷