/
githubmirror
/
interviews
Обзор
Документация
Войти
/
githubmirror
/
interviews
Код
Запросы
0
Пакеты
0
Релизы
0
Аналитика
Безопасность
master
leetcode/dynamic-programming/MinCostClimbingStairs.java
38 строк
1 KB
Kevin Naughton Jr
add MinCostClimbingStairs
29 май 2018, 20:20
29 май 2018, 20:20
84a1d4c
Код
Авторство
О чём код?
//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 with index 0, or the step with index 1. // //Example 1: //Input: cost = [10, 15, 20] //Output: 15 //Explanation: Cheapest is start on cost[1], pay that cost and go to the top. //Example 2: //Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1] //Output: 6 //Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3]. //Note: //cost will have a length in the range [2, 1000]. //Every cost[i] will be an integer in the range [0, 999]. class MinCostClimbingStairs { public int minCostClimbingStairs(int[] cost) { if(cost == null || cost.length == 0) { return 0; } if(cost.length == 1) { return cost[0]; } if(cost.length == 2) { return Math.min(cost[0], cost[1]); } int[] dp = new int[cost.length]; dp[0] = cost[0]; dp[1] = cost[1]; for(int i = 2; i < cost.length; i++) { dp[i] = Math.min(dp[i - 1] + cost[i], dp[i - 2] + cost[i]); } return Math.min(dp[cost.length - 1], dp[cost.length -2]); } }