题目描述

题解思路

我们使用a，b分别记录n - 2层向上爬的最小花费，n - 1层向上爬的最小花费
到达楼梯顶第N层，只能从N - 1层或者N - 2层向上爬
所以爬到第N层的最小花费 = 第N - 1层向上爬和第N - 2层向上爬的最小花费

题解代码

impl Solution {pub fn min_cost_climbing_stairs(cost: Vec<i32>) -> i32 {let (mut a, mut b) = (cost[0], cost[1]);for i in 2..cost.len() {(a, b) = (b, a.min(b) + cost[i]);}return a.min(b);}
}

题目链接

https://leetcode.cn/problems/min-cost-climbing-stairs/