【Day 54 】2023-04-08 - 746.使用最小花费爬楼梯

[azl397985856]

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

• 动态规划

  题目描述

  
  数组的每个下标作为一个阶梯，第 i 个阶梯对应着一个非负数的体力花费值 cost[i]（下标从 0 开始）。

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] 。

[Zoeyzyzyzy]

class Solution {
    // point: 到这一阶楼梯不收钱，只有从这个楼梯跳过去才收钱
    // Time Complexity: O(n), Space Complexity: O(n)
    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 0;
        for (int i = 2; i < n + 1; i++) {
            dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
        }
        return dp[n];
    }

    // 用滚动数组思想，将Space Complexity 减至O(1)
    public int minCostClimbingStairs1(int[] cost) {
        int n = cost.length;
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 0;
        for (int i = 2; i < n + 1; i++) {
            dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
        }
        return dp[n];
    }
}

[JiangyanLiNEU]

• DP
• T = O(N)
• S = O(1)

  class Solution(object):
  def minCostClimbingStairs(self, cost):
      if len(cost)==2: return min(cost)
      prev_two, prev_one = 0, min(cost[0], cost[1])
      for i in range(3, len(cost)):
          cur = min(prev_one+cost[i-1], prev_two+cost[i-2])
          prev_two, prev_one = prev_one, cur
      return min(prev_two+cost[-2], prev_one+cost[-1])

[SoSo1105]

class Solution(object): def minCostClimbingStairs(self, cost): if len(cost)==2: return min(cost) prev_two, prev_one = 0, min(cost[0], cost[1]) for i in range(3, len(cost)): cur = min(prev_one+cost[i-1], prev_two+cost[i-2]) prev_two, prev_one = prev_one, cur return min(prev_two+cost[-2], prev_one+cost[-1])

[NorthSeacoder]

/**
 * @param {number[]} cost
 * @return {number}
 */
var minCostClimbingStairs = function(cost) {
    let n = cost.length;
    //爬到 n 阶的最小花费
    const dp = [];
    dp[0] = cost[0];
    dp[1] = cost[1];
    for (let i = 2; i <= n; i++) {
        dp[i] = Math.min(dp[i - 1], dp[i - 2]) + (cost[i] || 0);
    }
    return dp.at(-1);
};

• TC:O(n)
• SC:O(n)

[Abby-xu]

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        if len(cost) < 2: return cost[0]

        dp = [0 for _ in range(len(cost) + 1)]
        dp[0], dp[1] = cost[0], cost[1]
        cost = cost + [0]

        for i in range(2, len(dp)): dp[i] = cost[i] + min(dp[i - 1], dp[i - 2])

        return dp[-1]

[Size-of]

/**

• @param {number[]} cost
• @return {number} */ var minCostClimbingStairs = function(cost) { const n = cost.length; const dp = new Array(n + 1); dp[0] = dp[1] = 0; for (let i = 2; i <= n; i++) { dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]); } return dp[n]; };

[csthaha]

var minCostClimbingStairs = function(cost) {
    const n = cost.length;
    const dp = new Array(n + 1);
    dp[0] = dp[1] = 0;
    for (let i = 2; i <= n; i++) {
        dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
    }
    return dp[n];
};

[harperz24]

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        tab = [0] * (len(cost) + 1)

        for i in range(2, len(cost) + 1):
            tab[i] = min(tab[i - 1] + cost[i - 1], tab[i - 2] + cost[i - 2])

        return tab[-1]

        # time: O(n)
        # space: O(n)

[Meisgithub]

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        vector<int> dp(n + 1, INT_MAX);
        dp[0] = 0, dp[1] = 0;
        for (int i = 2; i <= n; ++i)
        {
            dp[i] = min(dp[i - 2] + cost[i - 2], dp[i - 1] + cost[i - 1]);
        }
        return dp[n];
    }
};

[wangzh0114]

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        for (int i = 2; i < cost.size(); i++) {
            cost[i] = min(cost[i - 2], cost[i - 1]) + cost[i];
        }
        return min(cost[cost.size() - 2], cost[cost.size() - 1]);
    }
};

[FireHaoSky]

思路：动态规划，迭代计算没一层需要使用的步数

代码：python

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        pre = cur = 0
        for i in range(2, len(cost)+1):
            nxt = min(cur + cost[i - 1], pre + cost[i - 2])
            pre, cur = cur, nxt

        return cur

复杂度分析：

"""
时间复杂度：O(n)
空间复杂度：O(1)
"""

[bingzxy]

思路

动态规划 定义 dp[i] 为到台阶 i 所需花费的最小费用，状态转移方程：dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i];

代码

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int len = cost.length;
        int[] dp = new int[len + 1];
        dp[1] = cost[0];
        for (int i = 2; i <= len; i++) {
            dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i - 1];
        }
        return Math.min(dp[len - 1], dp[len]);
    }
}

复杂度分析

• 时间复杂度：O(n)
• 空间复杂度：O(n)

[Fuku-L]

代码

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];
    }
}

复杂度分析

• 时间复杂度：O(N)，其中 N 为数组长度。
• 空间复杂度：O(1)

[RestlessBreeze]

code

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        vector<int> dp(cost.size() + 1);
        dp[0] = 0;
        dp[1] = 0;
        for (int i = 2; i < dp.size(); i++)
            dp[i] = min(dp[i - 2] + cost[i - 2], dp[i - 1] + cost[i - 1]);
        return dp[cost.size()];
    }
};

[Diana21170648]

思路

动态规划

代码

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        dp = [0] * (len(cost)+1)
        dp[0], dp[1] = cost[0], cost[1]
        for i in range(2, len(cost)+1):
            dp[i] = min(dp[i-1], dp[i-2]) + (cost[i] if i != len(cost) else 0)
        return dp[-1]

**复杂度分析**
- 时间复杂度：O(N)，其中 N 为数组长度。
- 空间复杂度：O(N)

[yingchehu]

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        stair = [] 
        for i in range(len(cost)+1):
            if i == 0 or i == 1:
                stair.append(0)
            else:
                stair.append(min(stair[i-1]+cost[i-1], stair[i-2]+cost[i-2]))

        return stair[-1]

複雜度

Time: O(N) Space: O(N)

[zhangyu1131]

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        vector<int> dp(n, 0);
        for (int i = 2;  i < n; ++i)
        {
            dp[i] = std::min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
        }

        return std::min(dp[n - 1] + cost[n - 1], dp[n - 2] + cost[n - 2]);
    }
};

[JasonQiu]

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        if not cost:
            return 0            
        dp = [0] * len(cost)     
        dp[0] = cost[0]     
        if len(cost) >= 2:
            dp[1] = cost[1]
        for i in range(2, len(cost)):
            dp[i] = cost[i] + min(dp[i-1], dp[i-2])
        return min(dp[-1], dp[-2])

[kangliqi1]

class Solution { public int minCostClimbingStairs(int[] cost) {

    if (cost == null || cost.length == 0)
        return 0;

    int[] dp = new int[cost.length + 1];
    dp[0] = cost[0];
    dp[1] = cost[1];

    for(int i = 2; i <= cost.length; i++)
        dp[i] = Math.min(dp[i - 1], dp[i - 2]) + (i == cost.length ? 0 : cost[i]);

    return dp[cost.length];
}

[aoxiangw]

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        cost.append(0)
        for i in range(2, len(cost)):
            cost[i] += min(cost[i-2], cost[i-1])
        return cost[-1]

time, space: O(n), O(1)

[lp1506947671]

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        dp = [0] * (len(cost)+1)
        dp[0], dp[1] = cost[0], cost[1]
        for i in range(2, len(cost)+1):
            dp[i] = min(dp[i-1], dp[i-2]) + (cost[i] if i != len(cost) else 0)
        return dp[-1]

杂度分析

设：N阶台阶

• 时间复杂度：O(N)
• 空间复杂度：O(N)

[Lydia61]

746. 使用最小花费爬楼梯

代码

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]

复杂度分析

• 时间复杂度：O(N)
• 空间复杂度：O(N)

[snmyj]

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
       vector<int> dp(cost.size()+1,0);
       dp[0]=0;
       dp[1]=0;
       for(int i=2;i<=cost.size();i++){
           dp[i]=min(dp[i-2]+cost[i-2],dp[i-1]+cost[i-1]);

       }
        return dp[cost.size()];
    }

};

[jmaStella]

思路

dp

代码

public int minCostClimbingStairs(int[] cost) {
        int[] dp = new int[cost.length+1];

        for(int i=0; i<dp.length; i++){
            if(i<=1){
                dp[i] = 0;
                continue;
            }
            dp[i] = Math.min(dp[i-1]+cost[i-1], dp[i-2]+cost[i-2]);
        }
        return dp[dp.length-1];
    }

复杂度

时间 O(N) 空间 O(N)

[Jetery]

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        vector<int> dp(n);
        dp[0] = cost[0], dp[1] = cost[1];
        for (int i = 2; i < n; i++) {
            dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];
        }
        return min(dp[n - 1], dp[n - 2]);
    }
};

[joemonkeylee]

思路

dp

代码

        public int MinCostClimbingStairs(int[] cost)
        {
            int[] dp = new int[cost.Length + 1];
            dp[0] = 0;
            dp[1] = 0;
            for (int i = 2; i < dp.Length; i++)
            {
                dp[i] = Math.Min(dp[i - 2] + cost[i - 2], dp[i - 1] + cost[i - 1]);
            }
            return dp[cost.Length];
        }

复杂度分析

• 时间复杂度：O(n)
• 空间复杂度：O(n)

[bookyue]

TC: O(n)
SC: O(1)

    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;

        int secPrev = cost[0];
        int prev = cost[1];

        int ans;
        for (int i = 2; i < n; i++) {
            ans = Math.min(secPrev, prev) + cost[i];
            secPrev = prev;
            prev = ans;
        }

        return Math.min(prev, secPrev);
    }

[tzuikuo]

思路

动态规划

代码

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n=cost.size();
        vector<int> dp(n+1);
        if(n==1) return cost[0];
        if(n==2) return min(cost[0],cost[1]);
        for(int i=2;i<=n;i++){
            dp[i]=min(dp[i-1]+cost[i-1],dp[i-2]+cost[i-2]);
        }
        return dp[n];
    }
};

复杂度分析

• 时间复杂度：O(N)
• 空间复杂度：O(N)

[kofzhang]

思路

动态规划

复杂度

时间复杂度：O(n) 空间复杂度：O(n)

代码

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        l = len(cost)
        dp = [0]*l
        dp[0]=cost[0]
        dp[1]=cost[1]
        for i in range(2,l):
            dp[i]=min(dp[i-2],dp[i-1])+cost[i]
        return min(dp[-1],dp[-2])

[chanceyliu]

代码

function minCostClimbingStairs(cost: number[]): number {
  const n = cost.length;
  const dp = new Array(n + 1);
  dp[0] = dp[1] = 0;
  for (let i = 2; i <= n; i++) {
      dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
  }
  return dp[n];
};

复杂度分析

• 时间复杂度：O(n)
• 空间复杂度：O(n)

[CruiseYuGH]

思路

关键点

代码

• 语言支持：Python3

Python3 Code:

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        n = len(cost)+1
        dp = [0]*n
        dp[0]=0
        for i in range(2,n):
            dp[i]=min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2])
        return dp[-1]

复杂度分析

令 n 为数组长度。

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$