代码随想录day46|139. 单词拆分

139. 单词拆分

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):
            for j in wordDict:
                if i>=len(j) and (s[i-len(j):i] in wordDict) and dp[i-len(j)]:
                    dp[i] = True
        return dp[len(s)]

多重背包即为每个物品有并不是只能用一次，也不是无限重复使用，而是有nums数组控制使用物品的个数，将其展开就变成了0-1背包问题

def test_multi_pack():
    weight = [1, 3, 4]
    value = [15, 20, 30]
    nums = [2, 3, 2]
    bagWeight = 10
 
    # 将数量大于1的物品展开
    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):  # 遍历背包容量
            dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
        for j in range(bagWeight + 1):
            print(dp[j], end=" ")
        print()
 
    print(dp[bagWeight])
 
 
test_multi_pack()

问能否能装满背包（或者最多装多少）：dp[j] = max(dp[j], dp[j - nums[i]] + nums[i])

问装满背包有几种方法：dp[j] += dp[j - nums[i]]

问背包装满最大价值：dp[j] = max(dp[j], dp[j - weight[i]] + value[i])

问装满背包所有物品的最小个数：dp[j] = min(dp[j - coins[i]] + 1, dp[j])