【算法】动态规划：python实现 1

# 动态规划实现青蛙跳跃问题
def frog_jump(n: int) -> int:"""计算青蛙跳跃到第n级台阶的不同跳法数量:param n: 台阶数:return: 跳法数量"""if n <= 1:return 1dp = [0] * (n + 1)dp[0], dp[1] = 1, 1for i in range(2, n + 1):dp[i] = dp[i - 1] + dp[i - 2]return dp[n]# 动态规划实现爬楼梯
def climb_stairs(n: int) -> int:"""计算爬到第n级台阶的不同方法数量:param n: 台阶数:return: 方法数量"""if n <= 1:return 1dp = [0] * (n + 1)dp[0], dp[1] = 1, 1for i in range(2, n + 1):dp[i] = dp[i - 1] + dp[i - 2]return dp[n]# 动态规划实现最长递增子序列
def longest_increasing_subsequence(nums: list) -> int:"""计算给定数组的最长递增子序列长度:param nums: 输入数组:return: 最长递增子序列长度"""if not nums:return 0dp = [1] * len(nums)for i in range(len(nums)):for j in range(i):if nums[i] > nums[j]:dp[i] = max(dp[i], dp[j] + 1)return max(dp)#动态规划实现打家劫舍
def rob(nums: list) -> int:"""计算打家劫舍问题的最大收益:param nums: 每个房屋的金额列表:return: 最大收益"""if not nums:return 0if len(nums) == 1:return nums[0]dp = [0] * len(nums)dp[0], dp[1] = nums[0], max(nums[0], nums[1])for i in range(2, len(nums)):dp[i] = max(dp[i - 1], dp[i - 2] + nums[i])return dp[-1]#动态规划实现最小费用怕楼梯
def min_cost_climbing_stairs(cost: list) -> int:"""计算最小费用爬楼梯问题的最小费用:param cost: 每个台阶的费用列表:return: 最小费用"""if not cost:return 0n = len(cost)if n == 1:return cost[0]dp = [0] * (n + 1)dp[0], dp[1] = 0, cost[0]for i in range(2, n + 1):dp[i] = min(dp[i - 1] + (cost[i - 1] if i - 1 < n else 0),dp[i - 2] + (cost[i - 2] if i - 2 < n else 0))return dp[n]#动态规划实现最长公共子序列
def longest_common_subsequence(text1: str, text2: str) -> int:"""计算最长公共子序列的长度:param text1: 第一个字符串:param text2: 第二个字符串:return: 最长公共子序列的长度"""m, n = len(text1), len(text2)dp = [[0] * (n + 1) for _ in range(m + 1)]for i in range(1, m + 1):for j in range(1, n + 1):if text1[i - 1] == text2[j - 1]:dp[i][j] = dp[i - 1][j - 1] + 1else:dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])return dp[m][n]# 动态规划实现编辑距离
def edit_distance(str1: str, str2: str) -> int:"""计算两个字符串之间的编辑距离:param str1: 第一个字符串:param str2: 第二个字符串:return: 编辑距离"""m, n = len(str1), len(str2)dp = [[0] * (n + 1) for _ in range(m + 1)]for i in range(m + 1):for j in range(n + 1):if i == 0:dp[i][j] = j  # 如果str1为空，插入所有字符到str2elif j == 0:dp[i][j] = i  # 如果str2为空，删除所有字符从str1elif str1[i - 1] == str2[j - 1]:dp[i][j] = dp[i - 1][j - 1]  # 字符相同，不需要操作else:dp[i][j] = min(dp[i - 1][j] + 1,   # 删除dp[i][j - 1] + 1,   # 插入dp[i - 1][j - 1] + 1)  # 替换return dp[m][n]#动态规划实现第 N 个泰波那契数
def tribonacci(n: int) -> int:"""计算第 N 个泰波那契数:param n: N 的值:return: 第 N 个泰波那契数"""if n == 0:return 0elif n == 1 or n == 2:return 1dp = [0] * (n + 1)dp[0], dp[1], dp[2] = 0, 1, 1for i in range(3, n + 1):dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3]return dp[n]