On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step w…

问题描述: On a staircase, the i-th step has some non-negative cost cost[i]assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the s…

目录 题目链接 注意点 解法 小结 题目链接 Min Cost Climbing Stairs - LeetCode 注意点 注意边界条件 解法 解法一:这道题也是一道dp题.dp[i]表示爬到第i层的最小cost,想要到达第i层只有两种可能性,一个是从第i-2层上直接跳上来,一个是从第i-1层上跳上来.所以可以得到dp[i] = min(dp[i- 2] + cost[i - 2], dp[i - 1] + cost[i - 1]).时间复杂度O(n). class Solution { pu…

746. 使用最小花费爬楼梯 746. Min Cost Climbing Stairs 题目描述 数组的每个索引做为一个阶梯,第 i 个阶梯对应着一个非负数的体力花费值 cost[i].(索引从 0 开始) 每当你爬上一个阶梯你都要花费对应的体力花费值,然后你可以选择继续爬一个阶梯或者爬两个阶梯. 您需要找到达到楼层顶部的最低花费.在开始时,你可以选择从索引为 0 或 1 的元素作为初始阶梯. 每日一算法2019/5/14Day 11LeetCode746. Min Cost Climbing…

Leetcode之动态规划(DP)专题-746. 使用最小花费爬楼梯(Min Cost Climbing Stairs) 数组的每个索引做为一个阶梯,第 i个阶梯对应着一个非负数的体力花费值 cost[i](索引从0开始). 每当你爬上一个阶梯你都要花费对应的体力花费值,然后你可以选择继续爬一个阶梯或者爬两个阶梯. 您需要找到达到楼层顶部的最低花费.在开始时,你可以选择从索引为 0 或 1 的元素作为初始阶梯. 示例 1: 输入: cost = [10, 15, 20] 输出: 15 解释: 最…

leetcode 746. Min Cost Climbing Stairs(easy understanding dp solution) On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost…

lc 746 Min Cost Climbing Stairs 746 Min Cost Climbing Stairs On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach t…

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed). Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step w…

problem 746. Min Cost Climbing Stairs 题意: solution1:动态规划: 定义一个一维的dp数组,其中dp[i]表示爬到第i层的最小cost,然后来想dp[i]如何推导.思考一下如何才能到第i层呢?是不是只有两种可能性,一个是从第i-2层上直接跳上来,一个是从第i-1层上跳上来.不会再有别的方法,所以dp[i]只和前两层有关系,所以可以写做如下: dp[i] = min(dp[i- 2] + cost[i - 2], dp[i - 1] + cost[i…

Min Cost Climbing Stairs [746] 题目描述 简单来说就是:要跳过当前楼梯需要花费当前楼梯所代表的价值cost[i], 花费cost[i]之后,可以选择跳一阶或者两阶楼梯,以最小的代价达到楼层,也就是跨过所有楼梯 问题解决 穷举法 从第一阶楼梯开始,遍历所有可能的情况,然后选择代价最小的.复杂度会比较高,时间复杂度O(2^n),不太适合 动态规划 逆向解决:从后往前倒退,跳过当前阶楼梯代价最小的情况下需要考虑其面楼梯代价最小的情况,而且每次只能跳一阶或者两阶,也就是说跳…