LC 416 分割等和子集
class Solution:
def canPartition(self, nums: List[int]) -> bool:
target = sum(nums)
if target % 2 == 1:return False #总和为奇数,不能平分
target //= 2 #平分target
dp = [0] * 10001 #dp[j]表示 背包总容量(所能装的总重量)是j,放进物品后,背的最大重量为dp[j]
for i in range(len(nums)):
for j in range(target, nums[i] - 1, -1):
#如果使用一维dp数组,物品遍历的for循环放在外层,遍历背包的for循环放在内层,且内层for循环倒序遍历!
dp[j] = max(dp[j], dp[j - nums[i]] + nums[i])
#相当于背包里放入数值,那么物品i的重量是nums[i],其价值也是nums[i]
return target == dp[target] #如果dp[j] == j 说明,集合中的子集总和正好可以凑成总和j
01背包 二维数组表达:
def test_2_wei_bag_problem1(bag_size, weight, value) -> int:
rows, cols = len(weight), bag_size + 1
dp = [[0 for _ in range(cols)] for _ in range(rows)]
# 初始化dp数组.
for i in range(rows):
dp[i][0] = 0
first_item_weight, first_item_value = weight[0], value[0]
for j in range(1, cols):
if first_item_weight <= j:
dp[0][j] = first_item_value
# 更新dp数组: 先遍历物品, 再遍历背包.
for i in range(1, len(weight)):
cur_weight, cur_val = weight[i], value[i]
for j in range(1, cols):
if cur_weight > j: # 说明背包装不下当前物品.
dp[i][j] = dp[i - 1][j] # 所以不装当前物品.
else:
# 定义dp数组: dp[i][j] 前i个物品里,放进容量为j的背包,价值总和最大是多少。
dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - cur_weight]+ cur_val)
print(dp)
if __name__ == "__main__":
bag_size = 4
weight = [1, 3, 4]
value = [15, 20, 30]
test_2_wei_bag_problem1(bag_size, weight, value)
一维数组:
def test_1_wei_bag_problem():
weight = [1, 3, 4]
value = [15, 20, 30]
bag_weight = 4
# 初始化: 全为0
dp = [0] * (bag_weight + 1)
# 先遍历物品, 再遍历背包容量
for i in range(len(weight)):
for j in range(bag_weight, weight[i] - 1, -1):
# 递归公式
dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
print(dp)
test_1_wei_bag_problem()