＜代码随想录＞ 算法训练营-2024.12.20

322. 零钱兑换

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        # dp[i][j]表示 提供到coins[i]的硬币，总金额为j的最少硬币个数 硬币个数无限，完全背包
        # 有两种取值，一种是取dp[i-1][j] 另一种是如果j比当前面额小，则可以取dp[m][n-coins[m]]+1
        size = len(coins)
        dp = [[10**4 + 1] * (amount + 1) for _ in range(size)]

        for i in range(size):
            dp[i][0] = 0

        for m in range(size):
            for n in range(1, amount + 1):
                if coins[m] <= n:
                    dp[m][n] = min(dp[m - 1][n], dp[m][n - coins[m]] + 1)
                else:
                    dp[m][n] = dp[m - 1][n]

        return -1 if dp[-1][-1] == 10**4 + 1 else dp[-1][-1]

279. 完全平方数

class Solution:
    def numSquares(self, n: int) -> int:
        #预处理完全平方数数组
        # 题目变成，完全背包中达到目标值的最少元素个数
        nums = []
        temp = 1
        while temp**2 <= n:
            nums.append(temp**2)
            temp += 1
        if nums[-1] == n:
            return 1

        size = len(nums)
        dp = [[10**4 + 1] * (n + 1) for _ in range(size)]

        for i in range(size):
            dp[i][0] = 0

        for i in range(size):
            for j in range(1, n + 1):
                if nums[i] <= j:
                    dp[i][j] = min(dp[i - 1][j], dp[i][j - nums[i]] + 1)
                else:
                    dp[i][j] = dp[i - 1][j]

        return -1 if dp[-1][-1] == 10**4 + 1 else dp[-1][-1]

139. 单词拆分

class Solution(object):
    def wordBreak(self, s, wordDict):

        # 先对单词按长度排序
        wordDict.sort(key=lambda x: len(x))
        n = len(s)
        dp = [False] * (n + 1)
        dp[0] = True
        # 遍历背包
        for i in range(1, n + 1):
            # 遍历单词
            for word in wordDict:
                # 简单的 “剪枝”
                if len(word) > i:
                    break
                dp[i] = dp[i] or (dp[i - len(word)] and s[i - len(word): i] == word)
        return dp[-1]