动态规划详解（leetcode例题+解析）python

一级目录

例题+解析

1. 斐波那契数
   

class Solution:
    def fib(self, N: int) -> int:
        for i in range(N+1):
            if i == 0:
                res = 0
                pre2 = 0
            elif i == 1:
                res = 1
                pre1 = 1
            else:
                res = pre1+pre2
                pre1, pre2 = res, pre1
        return res

2. 零钱兑换

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

3. 零钱兑换II

5. 最长上升子序列

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        dp = {}
        for i in range(len(nums)):
            if i == 0:
                dp[i] = 1
            else:
                dp[i] = 1
                for j in range(i):
                    if nums[j] < nums[i]:
                        dp[i] = max(dp[i],dp[j]+1)
        res = 0
        for i in range(len(nums)):
            res = max(res,dp[i])
        return res

4. 连续子数组的最大和
   

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        dp = {}
        for i in range(len(nums)):
            if i == 0:
                dp[i] = nums[i]
            else:
                dp[i] = max(nums[i], dp[i-1]+nums[i])
        res = -float('inf')
        for i in range(len(nums)):
            if dp[i] > res:
                res = dp[i]
        return res

5. 背包问题（该题leetcode没有原题，但是有一些变形的题目）

6. 分割等和子集

class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        sum_all = sum(nums)
        if sum_all%2 != 0:
            return False
        sum_all = sum_all/2
        dp = [[False] * int(sum_all+1) for _ in range(len(nums)+1)]
        for i in range(len(nums)+1):
            dp[i][0]=True      
        for i in range(1, len(nums)+1):
            for j in range(1,int(sum_all)+1):
                if nums[i-1] > j:
                    dp[i][j] = dp[i-1][j]
                    continue
                if j ==  nums[i-1]:
                    dp[i][j] = True
                else:
                    dp[i][j] = dp[i-1][j] or dp[i-1][j-nums[i-1]]
        return dp[-1][-1]

压缩

class Solution:
    def canPartition(self, nums: List[int]) -> bool:
        sum_all = sum(nums)
        if sum_all%2 != 0:
            return False
        sum_all = sum_all/2
        dp = [False for _ in range(int(sum_all) + 1)]
        dp[0] = True
        for i in range(1, len(nums) + 1):
          for j in range(int(sum_all), 0,-1):
            if nums[i - 1] > j:
                dp[j] = dp[j]
                continue
            if j == nums[i - 1]:
                dp[j] = True
            else:
                dp[j] = dp[j] or dp[j - nums[i - 1]]
        return dp[-1]

三级目录

本文地址：https://blog.csdn.net/weixin_43494312/article/details/107159626