动态规划问题的一般形式：求最值、求最长递增子序列、最小编辑距离
动态规划问题一定具备最优子结构
重叠子问题、最优子结构、状态转移方程
定义状态转移方程：明确[状态]-->定义dp数组/函数的含义-->明确[选择]-->明确base case
1 def fib(n): 2 if n == 1 or n == 2: 3 return 1 4 return fib(n-1) + fib(n-2)
# 带备忘录的递归解法-->自顶向下的动态规划
1 def fib(n): 2 memo = [0 for i in range(n+1)] 3 return helper(memo, n) 4 5 def helper(memo, n): 6 if n==1 or n == 2: 7 return 1 8 if memo[n] != 0: 9 return memo[n] 10 memo[n] = helper(memo, n-1) + helper(memo, n-2) 11 return memo[n]
1 # 动态规划解法-->自底向上的迭代解法 2 def fib(n): 3 dp = [0 for i in range(n+1)] 4 dp[1] = dp[2] = 1 5 for i in range(3, n+1): 6 dp[i] = dp[i-1] + dp[i-2] 7 return dp[n]
1 # 优化 2 def fib(n): 3 dp1 = dp2 = 1 4 for i in range(3, n+1): 5 dp1, dp2 = dp2, dp1+dp2 6 return dp2
凑零钱问题
1 # 递归 2 def coinChange(coins, amount): 3 # 定义：要凑出金额n，至少要dp(n)个硬币 4 def dp(n): 5 # base case 6 if n == 0: return 0 7 if n < 0: return -1 8 # 做选择，需要硬币最少的那个结果就是答案 9 res = float('inf') 10 for coin in coins: 11 subproblem = dp(n - coin) 12 if subproblem == -1: 13 continue 14 res = min(res, 1 + subproblem) 15 return res if res != float('inf') else -1 16 return dp[amount]
1 def coinChange(coins, amount): 2 memo = {} 3 def dp(n): 4 if n in memo: 5 return memo[n] 6 if n == 0: 7 return 0 8 if n < 0: 9 return -1 10 res = float('inf') 11 for coin in coins: 12 subproblem = dp(n-cpin) 13 if subproblem == -1: 14 continue 15 res = min(res, 1 + subproblem) 16 17 memo[n] = res if res != float('inf') else -1 18 return dp(amount)
1 def coinChange(coins, amount): 2 dp = [amount + 1 for i in range(amount+1)] 3 for i in range(amount+1): 4 for coin in coins: 5 if i - coin < 0: 6 continue 7 dp[i] = min(dp[i], 1 + dp[i - coin]) 8 return -1 if dp[amount] == amount+1 else dp[amount]