代码随想录刷题第四十三天
今天为三道0-1背包问题的变种， 分别有三个小问题
- 给定一个容量为j的背包，尽可能装下物品，找到能装下物品的最大价值
- dp[i][j] = max(dp[i-1][j], dp[i-1][j-nums[i]]+nums[i])
- 给定一个容量为j的背包，找到有多少种方法能够装满背包
- dp[i][j] += dp[i-1][j-nums[i]]
- 将左上角dp[0][0]初始化为1， 从这一种方法开始往上累加
- 给定一个容量为j的背包，找到背包最后装了多少个物品
- dp[i][j] = max(dp[i-1][j], dp[i-1][j-nums[i]]+1)
最后一块石头的重量II (LC 1049)
题目思路：
代码实现：
class Solution:def lastStoneWeightII(self, stones: List[int]) -> int:allsum = sum(stones)target = allsum//2dp = [[0 for _ in range(target+1)] for _ in range(len(stones))]if stones[0] <= target:for j in range(stones[0], target+1):dp[0][j] = stones[0]for i in range(1, len(stones)):for j in range (1, target+1):if stones[i] > j:dp[i][j] = dp[i-1][j]else:dp[i][j] = max(dp[i-1][j], dp[i-1][j-stones[i]]+stones[i])return allsum-dp[len(stones)-1][target]*2
目标和 (LC 494)
题目思路：
代码实现：
class Solution:def findTargetSumWays(self, nums: List[int], target: int) -> int:allsum =sum(nums)if (allsum+target)%2==1:return 0if allsum < abs(target):return 0 addtarget = (allsum+target)//2dp = [[0 for i in range(addtarget+1)] for _ in range(len(nums))]if nums[0]<=addtarget:dp[0][nums[0]] = 1 if nums[0]!=0:dp[0][0] = 1else:dp[0][0] = 2 for i in range(1, len(nums)):for j in range(addtarget+1):dp[i][j] = dp[i - 1][j] # 不选取当前元素if nums[i] <= j:dp[i][j] += dp[i-1][j-nums[i]]print(dp)return dp[len(nums)-1][addtarget]
一和零 (LC 474)
题目思路：
代码实现：
三维dp，会超时
class Solution:def findMaxForm(self, strs: List[str], m: int, n: int) -> int:dp = [[[0 for _ in range(n+1)] for _ in range(m+1) ] for _ in range(len(strs))]zeros, ones = self.count_zeros(strs[0])if zeros<=m and ones<=n:for j in range(zeros, m+1):for k in range(ones, n+1):dp[0][j][k] = 1 for i in range(1, len(strs)):for j in range(m+1):for k in range(n+1):zeros, ones = self.count_zeros(strs[i])dp[i][j][k] = dp[i-1][j][k] # 不选取当前元素if zeros<=j and ones<=k:dp[i][j][k] = max(dp[i-1][j][k], dp[i-1][j-zeros][k-ones]+1)print(dp)return dp[len(strs)-1][m][n]def count_zeros(self, input_string):count = 0for char in input_string:if char == '0':count += 1return count, len(input_string)-count
二维dp
class Solution:def findMaxForm(self, strs: List[str], m: int, n: int) -> int:dp = [[0 for _ in range(n+1)] for _ in range(m+1)]for i in range(len(strs)):for j in range(m, -1, -1):for k in range(n, -1, -1):zeros, ones = self.count_zeros(strs[i])dp[j][k] = dp[j][k] # 不选取当前元素if zeros<=j and ones<=k:dp[j][k] = max(dp[j][k], dp[j-zeros][k-ones]+1)return dp[m][n]def count_zeros(self, input_string):count = 0for char in input_string:if char == '0':count += 1return count, len(input_string)-count
