目录

学习目标

学习内容

 1049. 最后一块石头的重量 II 

  494. 目标和 

  474.一和零 

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学习目标

学习内容

 1049. 最后一块石头的重量 II 

1049. 最后一块石头的重量 II - 力扣（LeetCode）https://leetcode.cn/problems/last-stone-weight-ii/

class Solution:
    def lastStoneWeightII(self, stones: List[int]) -> int:
        amount = sum(stones)
        target = amount//2
        n = len(stones)
        dp = [0]*(target+1)
        for i in range(n):
            for j in range(target,stones[i]-1,-1):
                dp[j] = max(dp[j],dp[j-stones[i]]+stones[i])
        #print(dp)
        return amount-2*dp[-1]

  494. 目标和 

494. 目标和 - 力扣（LeetCode）https://leetcode.cn/problems/target-sum/

class Solution:
    def findTargetSumWays(self, nums: List[int], target: int) -> int:
        amount = sum(nums)
        if (amount-target)%2!=0 or amount<abs(target):return 0
        n = len(nums)
        target = (amount-target)//2
        dp = [0]*(target+1)
        dp[0] = 1
        for i in range(n):
            for j in range(target,nums[i]-1,-1):
                dp[j] +=dp[j-nums[i]]
        #print(dp)
        return dp[target]

  474.一和零 

474. 一和零 - 力扣（LeetCode）https://leetcode.cn/problems/ones-and-zeroes/

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        # dp[k][i][j] 表示 用到个k个元素当前装满i个0和j个1的最长子集长度
        # dp[k][i][j] = max(dp[k-1][i][j],dp[k-1][i-strs[i].count(0)][j-strs[i].count(1)]+1)
        # 初始化 dp[i][j]根据第一个元素进行初始化
        # 顺序 先物品 后背包 从上到下 从右到左（为了元素只用了一次）

        dp = [[0]*(n+1)for _ in range(m+1)]
        for item in strs:
            num0 = 0
            num1 = 0
            for c in item:
                if c=="0":
                    num0+=1
                else:
                    num1+=1
            for i in range(m,num0-1,-1):
                for j in range(n,num1-1,-1):
                    dp[i][j] = max(dp[i][j],dp[i-num0][j-num1]+1)
        return dp[-1][-1]