@@ -67,14 +67,15 @@ $dp[i][w] = \begin{cases} dp[i - 1][w] & w < weight[i - 1] \cr max \lbrace dp[i
6767
6868``` Python
6969class Solution :
70+ # 思路 1:动态规划 + 二维基本思路
7071 def zeroOnePackMethod1 (self weight : [int ], value : [int ], W : int ):
7172 size = len (weight)
7273 dp = [[0 for _ in range (W + 1 )] for _ in range (size + 1 )]
7374
7475 # 枚举前 i 种物品
7576 for i in range (1 , size + 1 ):
7677 # 枚举背包装载重量
77- for w in range (1 , W + 1 ):
78+ for w in range (W + 1 ):
7879 # 第 i - 1 件物品装不下
7980 if w < weight[i - 1 ]:
8081 # dp[i][w] 取「前 i - 1 件物品装入载重为 w 的背包中的最大价值」
@@ -137,6 +138,7 @@ $dp[w] = \begin{cases} dp[w] & w < weight[i - 1] \cr max \lbrace dp[w], dp[w - w
137138
138139``` Python
139140class Solution :
141+ # 思路 2:动态规划 + 滚动数组优化
140142 def zeroOnePackMethod2 (self weight : [int ], value : [int ], W : int ):
141143 size = len (weight)
142144 dp = [0 for _ in range (W + 1 )]
@@ -232,14 +234,18 @@ $dp[w] = \begin{cases} dp[w] & w < nums[i - 1] \cr max \lbrace dp[w], \quad dp[w
232234
233235``` Python
234236class Solution :
237+ # 思路 2:动态规划 + 滚动数组优化
235238 def zeroOnePackMethod2 (self weight : [int ], value : [int ], W : int ):
236239 size = len (weight)
237240 dp = [0 for _ in range (W + 1 )]
238241
242+ # 枚举前 i 种物品
239243 for i in range (1 , size + 1 ):
244+ # 逆序枚举背包装载重量(避免状态值错误)
240245 for w in range (W, weight[i - 1 ] - 1 , - 1 ):
246+ # dp[w] 取「前 i - 1 件物品装入载重为 w 的背包中的最大价值」与「前 i - 1 件物品装入载重为 w - weight[i - 1] 的背包中,再装入第 i - 1 物品所得的最大价值」两者中的最大值
241247 dp[w] = max (dp[w], dp[w - weight[i - 1 ]] + value[i - 1 ])
242- 248+
243249 return dp[W]
244250
245251 def canPartition (self nums : List[int ]) -> bool :
@@ -248,7 +254,7 @@ class Solution:
248254 return False
249255
250256 target = sum_nums // 2
251- return self .zeroOnePackOptimization (nums, nums, target) == target
257+ return self .zeroOnePackMethod2 (nums, nums, target) == target
252258```
253259
254260##### 思路 1:复杂度分析
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