Leetcode| 139. 单词拆分、多重背包 Day46

139. Word Break

dp[i] : 字符串长度为i的话，dp[i]为true，表示可以拆分为一个或多个在字典中出现的单词
如果([j, i] 这个区间的子串出现在字典里 && dp[j]是true) 那么 dp[i] = true

本题需要考虑顺序，所以先遍历背包，再遍历物品

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [False] * (len(s) + 1)
        dp[0] = True

        for i in range(len(s) + 1):     # 遍历背包
            for j in range(0, i):       # 遍历物品
                subStr = s[j: i]
                if subStr in wordDict and dp[j] == True:
                    dp[i] = True

        return dp[len(s)]

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多重背包

多重背包

多重背包每件物品最多有Mi件可用，把Mi件摊开，其实就是一个01背包问题了。

def test_multi_pack():
    weight = [1, 3, 4]
    value = [15, 20, 30]
    nums = [2, 3, 2]
    bagWeight = 10

    # 转换为0-1背包
    for i in range(len(nums)):
        while nums[i] > 1:
            weight.append(weight[i])
            value.append(value[i])
            nums[i] -= 1

    
    dp = [0] * (bagWeight + 1)
    for i in range(len(weight)):    # 遍历物品
        for j in range(bagWeight, weight[i] - 1, -1):   # 遍历背包
            dp[j] = max(dp[j], dp[j - weight[i]] + value[i])

    print(dp[bagWeight])

test_multi_pack()