Open azl397985856 opened 1 year ago
class Solution {
public int minCostClimbingStairs(int[] cost) {
int n = cost.length;
int[] dp = new int[n+1];
dp[0] = dp[1] = 0;
for(int i = 2; i<=n; i++){
dp[i] = Math.min(dp[i-1]+cost[i-1], dp[i-2] + cost[i-2]);
}
return dp[n];
}
}
思路:
这个问题可以通过动态规划来解决。我们可以从顶部开始向下爬楼梯,对于每一步,我们都计算到达这一步的最小成本。这个成本是当前步骤的成本加上到达下一步或下下一步的最小成本。最后,我们返回到达第0步或第1步的最小成本,因为我们可以从这两步开始爬楼梯。
代码:
class Solution:
def minCostClimbingStairs(self, cost: List[int]) -> int:
n = len(cost)
dp = [0] * (n + 1)
dp[n-1] = cost[n-1]
for i in range(n-2, -1, -1):
dp[i] = cost[i] + min(dp[i+1], dp[i+2])
return min(dp[0], dp[1])
复杂度分析:
时间复杂度:O(n) 空间复杂度:O(n)
class Solution {
func minCostClimbingStairs(_ cost: [Int]) -> Int {
if cost.isEmpty {
return 0
}
var dp = Array(repeating: 0, count: cost.count + 1)
dp[0] = cost[0]
dp[1] = cost[1]
for i in 2...cost.count {
dp[i] = min(dp[i - 1], dp[i - 2]) + (i == cost.count ? 0 : cost[i])
}
return dp[cost.count]
}
}
class Solution:
def minCostClimbingStairs(self, cost: List[int]) -> int:
minimum_cost = [0] * (len(cost) + 1)
for i in range(2, len(cost) + 1):
take_one_step = minimum_cost[i - 1] + cost[i - 1]
take_two_steps = minimum_cost[i - 2] + cost[i - 2]
minimum_cost[i] = min(take_one_step, take_two_steps)
return minimum_cost[-1]
动态规划
#20230802 第121(54)天 11:18
#T=O(n)
#S=O(1),滚动数组优化
class Solution:
def minCostClimbingStairs(self, cost: List[int]) -> int:
prev, curr = cost[0], cost[1]
for i in range(2, len(cost)+1):
next = min(prev, curr) + (cost[i] if i != len(cost) else 0)
prev, curr = curr, next
return curr
复杂度分析
class Solution: def minCostClimbingStairs(self, cost: List[int]) -> int: prev, curr = cost[0], cost[1] for i in range(2, len(cost)+1): next = min(prev, curr) + (cost[i] if i != len(cost) else 0) prev, curr = curr, next return curr
class Solution(object):
# dp
# 长度比cost长度多一个,用来放最初的数值0
# 后面的关键就在于取前两个dp的值加上对应cost的值,取小生成新DP值
# O(N), O(N)
def minCostClimbingStairs(self, cost):
if len(cost) < 2: return 0
if len(cost) == 2: return min(cost[0],cost[1])
dp = [0, 0]
for i in range(2, len(cost)+1):
dp.append(min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2]))
return dp[-1]
class Solution:
def minCostClimbingStairs(self, cost):
n = len(cost)
if n < 2:
return 0
dp = [0] * (n + 1) # 使用dp数组来保存截止每个阶梯的最低花费值
for i in range(2, n + 1):
dp[i] = min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2])
return dp[n]
ar minCostClimbingStairs = function(cost) { const n = cost.length; const dp = new Array(n);
dp[0] = cost[0]; dp[1] = cost[1];
for (let i = 2; i < n; i++) { dp[i] = Math.min(dp[i-1] + cost[i], dp[i-2] + cost[i]); }
return Math.min(dp[n-1], dp[n-2]); };
dp[i] = Math.min(dp[i-2] + cost[i-2], dp[i-1] + cost[i-1])
var minCostClimbingStairs = function (cost) {
let len = cost.length;
let dp = Array(len + 1).fill(0);
dp[0] = dp[1] = 0;
for (let i = 2; i <= len; i++) {
dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
}
console.log(dp);
return dp[len];
};
复杂度分析
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int n=cost.size();
vector<int> dp(n+1);
for(int i=2;i<n+1;i++){
dp[i]=min(dp[i-2]+cost[i-2],dp[i-1]+cost[i-1]);
}
return dp[n];
}
};
class Solution:
def minCostClimbingStairs(self, cost: List[int]) -> int:
n = len(cost)
minCost = [0] * n
minCost[1] = min(cost[0], cost[1])
for i in range(2, n):
minCost[i] = min(minCost[i - 1] + cost[i], minCost[i - 2] + cost[i - 1])
return minCost[-1]
746.使用最小花费爬楼梯
入选理由
暂无
题目地址
https://leetcode-cn.com/problems/min-cost-climbing-stairs/
前置知识
题目描述
每当你爬上一个阶梯你都要花费对应的体力值,一旦支付了相应的体力值,你就可以选择向上爬一个阶梯或者爬两个阶梯。
请你找出达到楼层顶部的最低花费。在开始时,你可以选择从下标为 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:最低花费是从 cost[1] 开始,然后走两步即可到阶梯顶,一共花费 15 。 示例 2:
输入:cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1] 输出:6 解释:最低花费方式是从 cost[0] 开始,逐个经过那些 1 ,跳过 cost[3] ,一共花费 6 。
提示:
cost 的长度范围是 [2, 1000]。 cost[i] 将会是一个整型数据,范围为 [0, 999] 。