From a275c2cc553ad35e50d736e0630947a5bdf2d13e Mon Sep 17 00:00:00 2001
From: YukunJ <yukunj@andrew.cmu.edu>
Date: Tue, 30 Aug 2022 21:51:30 -0400
Subject: [PATCH] =?UTF-8?q?Update:=20=E6=B3=A8=E9=87=8A=E5=A2=9E=E5=8A=A0?=
 =?UTF-8?q?=E4=BD=BF=E7=94=A8STL=E6=B1=82=E5=92=8C=E7=9A=84=E7=AE=80?=
 =?UTF-8?q?=E4=BE=BF=E5=B0=8F=E8=B4=B4=E5=A3=AB?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0416.分割等和子集.md | 9 +++++++--
 1 file changed, 7 insertions(+), 2 deletions(-)

diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md
index 9c90ea27..fa677da0 100644
--- a/problems/0416.分割等和子集.md
+++ b/problems/0416.分割等和子集.md
@@ -143,12 +143,17 @@ dp[j]的数值一定是小于等于j的。
 class Solution {
 public:
     bool canPartition(vector<int>& nums) {
-        // 使用标准库函数 便捷求和
-        int sum = accumulate(nums.begin(), nums.end(), 0);
+        int 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
         vector<int> dp(10001, 0);
+        for (int i = 0; i < nums.size(); i++) {
+            sum += nums[i];
+        }
+        // 也可以使用库函数一步求和
+        // int sum = accumulate(nums.begin(), nums.end(), 0);
         if (sum % 2 == 1) return false;
         int target = sum / 2;