递推算法刷题

70. 爬楼梯

class Solution:
    def climbStairs(self, n: int) -> int:
        a , b = 1, 1
        for i in range(2, n + 1):
            b, a = a + b, b
        return b 

746. 使用最小花费爬楼梯

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

256. 粉刷房⼦

假如有⼀排房⼦，共 n 个，每个房⼦可以被粉刷成红⾊、蓝⾊或者绿⾊这三种颜⾊中 的⼀种，你需要粉刷所有的房⼦并且使其相邻的两个房⼦颜⾊不能相同。 当然，因为市场上不同颜⾊油漆的价格不同，所以房⼦粉刷成不同颜⾊的花费成本也 是不同的。每个房⼦粉刷成不同颜⾊的花费是以⼀个 n x 3 的正整数矩阵 costs 来表 ⽰的。 例如， costs[0][0] 表⽰第 0 号房⼦粉刷成红⾊的成本花费； costs[1][2] 表⽰第 1 号 房⼦粉刷成绿⾊的花费，以此类推。 请计算出粉刷完所有房⼦最少的花费成本。

120. 三角形最小路径和

class Solution:
    def minimumTotal(self, triangle: List[List[int]]) -> int:
        n = len(triangle)
        dp = [[0 for _ in range(n)], [0 for _ in range(n)]]
        dp[(n - 1) % 2] = triangle[n - 1][:]
        for i in range(n-2, -1, -1):
            idx = i % 2
            pre_idx = 1 - idx
            for j in range(0,i + 1):
                dp[idx][j] = min(dp[pre_idx][j], dp[pre_idx][j + 1]) + triangle[i][j]
        
        return dp[0][0]

119. 杨辉三角 II

class Solution:
    def getRow(self, rowIndex: int) -> List[int]:
        n = rowIndex + 1
        f = [[1 for _ in range(n)], [1 for _ in range(n)]]
        for i in range(2, n):
            idx = i %  2
            pre_index =  1 - idx
            for j in range(1, i):
                f[idx][j] = f[pre_index][j] + f[pre_index][j-1]
        
        return f[rowIndex % 2]

53. 最⼤⼦序和

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        n = len(nums)
        for i in range(1, n): nums[i] += nums[i-1]
        pre = 0
        ans = float("-inf")
        for num in nums:
            ans = max(ans, num - pre)
            pre = min(pre, num)
        return ans 

122. 买卖股票的最佳时机 II

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        ans = 0
        pre = prices[0]
        for price in prices:
            if price > pre:
                ans += price - pre
            pre = price
        return ans 

198. 打家劫舍

class Solution:
    def rob(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [[0 for _ in range(2)] for _ in range(2)]
        dp[0][1] = nums[0]
        for i in range(1, n):
            idx = i % 2
            pre_idx = 1 - idx
            dp[idx][0] = max(dp[pre_idx][0], dp[pre_idx][1])
            dp[idx][1] =dp[pre_idx][0] + nums[i]
        return max(dp[(n - 1) % 2])

152. 乘积最大子数组

class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        pre_min = pre_max = 1 
        ans = float("-inf")
        for num in nums:
            if num < 0: pre_max, pre_min = pre_min, pre_max
            pre_max = max(pre_max * num, num)
            pre_min = min(pre_min * num, num)
            ans = max(pre_max, ans)
        return ans

322. 零钱兑换

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        dp = [-1] * (amount + 1)
        dp[0] = 0
        for coin in coins:
            for i in range(coin, amount + 1):
                if dp[i - coin] == -1: continue
                if dp[i] == -1 or dp[i] > dp[i - coin] + 1: dp[i] = dp[i - coin] + 1
                dp[i] = min(dp[i], dp[i-coin] + 1)
        return dp[amount]

300. 最长递增子序列

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        dp = []
        for i in range(len(nums)):
            if not dp or nums[i] > dp[-1]:
                dp.append(nums[i])
            else:
                l = 0
                r = len(dp) - 1
                tmp = 0
                while l <= r:
                    mid = (l + r) // 2
                    if dp[mid] >= nums[i]:
                        tmp = mid
                        r = mid - 1
                    else: l = mid + 1
                dp[tmp] = nums[i]
        return len(dp)