代码随想录day43|1049. 最后一块石头的重量 II494. 目标和474. 一和零

1049. 最后一块石头的重量 II

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

494. 目标和

class Solution:
    def findTargetSumWays(self, nums: List[int], target: int) -> int:
        # 想象成正组和负组pos -neg = target pos+neg = sum
        #dp 数组是正组和为pos的个数
        #递推公式是 dp[j] += dp[j-nums[i]]
        if abs(target) > sum(nums):
            return 0  # 此时没有方案
        if (target + sum(nums)) % 2 == 1:
            return 0  # 此时没有方案
        pos = int((sum(nums) + target)/2)
        dp = [0]* (pos+1)
        dp[0] =1
        for i in nums:
            for j in range(pos,-1,-1):
                if j>=i:
                    dp[j] +=dp[j-i]
            # print(dp)
        return dp[pos]

474. 一和零

class Solution:
    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
        dp = [[0] *(n+1) for _ in range(m+1)]
        for s in strs:
            nums0=s.count('0')
            nums1 = len(s)-nums0
            print(nums0)
            for i in range(m,nums0-1,-1):
                for j in range(n,nums1-1,-1):
                    dp[i][j] = max(dp[i][j],dp[i-nums0][j-nums1]+1)
        return dp[m][n]