动态规划5部曲

1. 确定dp数组（dp table）以及下标的含义
2. 确定递推公式
3. dp数组如何初始化
4. 确定遍历顺序
5. 举例推导dp数组

1049.最后一块石头的重量II

leetcode 1049.最后一块石头的重量II
代码随想录

class Solution:def lastStoneWeightII(self, stones: List[int]) -> int:stone_sum = sum(stones)target = stone_sum // 2dp = [0] * (target + 1)for stone in stones:for j in range(target, stone - 1, -1):# dp[j- stone] + stone # 表示把当前石头放在背包里，剩下的容量取最大价值dp[j-stone]dp[j] = max(dp[j], dp[j-stone] + stone)return stone_sum - dp[-1] - dp[-1]

494.目标和

leetcode 494.目标和
代码随想录

把直接求解转化成了求选出和为target的组合总数。

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

474.一和零

leetcode 474.一和零
代码随想录

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:zero_count_s = s.count("0")one_count_s = s.count("1")for i in range(m, zero_count_s - 1, -1):for j in range(n, one_count_s - 1, -1):dp[i][j] = max(dp[i][j], dp[i-zero_count_s][j-one_count_s] + 1)return dp[m][n]