LeetCode每日一题(2110. Number of Smooth Descent Periods of a Stock)-优快云博客

该博客讨论了一种计算股票价格序列中滑降期数量的方法。滑降期是指连续几天股票价格逐日下降且降幅为1的时期。给定一个整数数组表示股票历史价格，任务是找出所有滑降期。博客提供了一个实现方案，通过动态规划计算了以每个价格为终点的滑降期数量，并返回它们的总和。

You are given an integer array prices representing the daily price history of a stock, where prices[i] is the stock price on the ith day.

A smooth descent period of a stock consists of one or more contiguous days such that the price on each day is lower than the price on the preceding day by exactly 1. The first day of the period is exempted from this rule.

Return the number of smooth descent periods.

Example 1:

Input: prices = [3,2,1,4]
Output: 7

Explanation: There are 7 smooth descent periods:
[3], [2], [1], [4], [3,2], [2,1], and [3,2,1]
Note that a period with one day is a smooth descent period by the definition.

Example 2:

Input: prices = [8,6,7,7]
Output: 4

Explanation: There are 4 smooth descent periods: [8], [6], [7], and [7]
Note that [8,6] is not a smooth descent period as 8 - 6 ≠ 1.

Example 3:

Input: prices = [1]
Output: 1
Explanation: There is 1 smooth descent period: [1]

Constraints:

1 <= prices.length <= 105
1 <= prices[i] <= 105

假设 dp[i]是到 prices[i]为终点的 smooth descent periods 的数量，那如果 prices[i] - prices[i+1] == 1, 则 dp[i+1] = dp[i] + 1, 最终的结果是 sum(dp[0], dp[1], …, dp[n])

impl Solution {
    pub fn get_descent_periods(prices: Vec<i32>) -> i64 {
        let mut dp = vec![1i64; prices.len()];
        for i in 1..prices.len() {
            if prices[i - 1] - prices[i] == 1 {
                dp[i] += dp[i - 1]
            }
        }
        dp.into_iter().sum()
    }
}