From 5ab0a94c7152431b23303a6e64dc18a07c9ed37f Mon Sep 17 00:00:00 2001
From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com>
Date: Sun, 4 Jun 2023 05:36:45 -0500
Subject: [PATCH] =?UTF-8?q?Update=200416.=E5=88=86=E5=89=B2=E7=AD=89?=
 =?UTF-8?q?=E5=92=8C=E5=AD=90=E9=9B=86.md?=
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---
 problems/0416.分割等和子集.md | 107 +++++++++++++++++-----------
 1 file changed, 65 insertions(+), 42 deletions(-)

diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md
index 8115e18e..bb210e29 100644
--- a/problems/0416.分割等和子集.md
+++ b/problems/0416.分割等和子集.md
@@ -296,62 +296,85 @@ false true false false false true true false false false true true
 ```
 
 ### Python：
+卡哥版
 ```python
-# 一维度数组解法
 class Solution:
     def canPartition(self, nums: List[int]) -> bool:
-        target = sum(nums)
-        if target % 2 == 1: return False
-        target //= 2
-        dp = [0] * (target + 1)
-        for i in range(len(nums)):
-            for j in range(target, nums[i] - 1, -1):
-                dp[j] = max(dp[j], dp[j - nums[i]] + nums[i])
-        return target == dp[target]
+        _sum = 0
+
+        # dp[i]中的i表示背包内总和
+        # 题目中说：每个数组中的元素不会超过 100，数组的大小不会超过 200
+        # 总和不会大于20000，背包最大只需要其中一半，所以10001大小就可以了
+        dp = [0] * 10001
+        for num in nums:
+            _sum += num
+        # 也可以使用内置函数一步求和
+        # _sum = sum(nums)
+        if _sum % 2 == 1:
+            return False
+        target = _sum // 2
+
+        # 开始 0-1背包
+        for num in nums:
+            for j in range(target, num - 1, -1):  # 每一个元素一定是不可重复放入，所以从大到小遍历
+                dp[j] = max(dp[j], dp[j - num] + num)
+
+        # 集合中的元素正好可以凑成总和target
+        if dp[target] == target:
+            return True
+        return False
+
 ```
-
+二维DP版
 ```python
-# 二维度数组解法
 class Solution:
     def canPartition(self, nums: List[int]) -> bool:
-        target = sum(nums)
-        nums = sorted(nums)
+        
+        total_sum = sum(nums)
 
-        # 做最初的判断
-        if target % 2 != 0:
+        if total_sum % 2 != 0:
             return False
 
-        # 找到 target value 可以认为这个是背包的体积
-        target = target // 2
+        target_sum = total_sum // 2
+        dp = [[False] * (target_sum + 1) for _ in range(len(nums) + 1)]
 
-        row = len(nums)
-        col = target + 1
+        # 初始化第一行（空子集可以得到和为0）
+        for i in range(len(nums) + 1):
+            dp[i][0] = True
 
-        # 定义 dp table
-        dp = [[0 for _ in range(col)] for _ in range(row)] 
-
-        # 初始 dp value
-        for i in range(row):
-            dp[i][0] = 0
-        
-        for j in range(1, target):
-            if nums[0] <= j:
-                dp[0][j] = nums[0]
-
-        # 遍历 先遍历物品再遍历背包
-        for i in range(1, row):
-
-            cur_weight = nums[i]
-            cur_value = nums[i]
-
-            for j in range(1, col):
-                if cur_weight > j:
+        for i in range(1, len(nums) + 1):
+            for j in range(1, target_sum + 1):
+                if j < nums[i - 1]:
+                    # 当前数字大于目标和时，无法使用该数字
                     dp[i][j] = dp[i - 1][j]
                 else:
-                    dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - cur_weight] + cur_value)
-            
-        # 输出结果
-        return dp[-1][col - 1] == target   
+                    # 当前数字小于等于目标和时，可以选择使用或不使用该数字
+                    dp[i][j] = dp[i - 1][j] or dp[i - 1][j - nums[i - 1]]
+
+        return dp[len(nums)][target_sum]
+
+```
+一维DP版
+```python
+class Solution:
+    def canPartition(self, nums: List[int]) -> bool:
+
+        total_sum = sum(nums)
+
+        if total_sum % 2 != 0:
+            return False
+
+        target_sum = total_sum // 2
+        dp = [False] * (target_sum + 1)
+        dp[0] = True
+
+        for num in nums:
+            # 从target_sum逆序迭代到num，步长为-1
+            for i in range(target_sum, num - 1, -1):
+                dp[i] = dp[i] or dp[i - num]
+
+        return dp[target_sum]
+ 
 ```
 
 ### Go：