From 8a335ef25af8585f0b2428c8b568e9e7655f8f13 Mon Sep 17 00:00:00 2001
From: markilue <745518019@qq.com>
Date: Fri, 25 Nov 2022 22:09:23 +0800
Subject: [PATCH] =?UTF-8?q?leecode=E6=9B=B4=E6=96=B0?=
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---
 .../com/markilue/leecode/dynamic/T10_FindTargetSumWays.java     | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/Leecode/src/main/java/com/markilue/leecode/dynamic/T10_FindTargetSumWays.java b/Leecode/src/main/java/com/markilue/leecode/dynamic/T10_FindTargetSumWays.java
index 0d2e6ac..db850cb 100644
--- a/Leecode/src/main/java/com/markilue/leecode/dynamic/T10_FindTargetSumWays.java
+++ b/Leecode/src/main/java/com/markilue/leecode/dynamic/T10_FindTargetSumWays.java
@@ -30,7 +30,7 @@ public class T10_FindTargetSumWays {
      *                  已经有一个5 （nums[i]）的话，有 dp[0]中方法 凑成 dp[5]
      *                  所以求组合类问题的公式，都是类似这种：
      *                      dp[j] += dp[j - nums[i]]
-     *          3)dp数组初始化:  dp[i][0]=0
+     *          3)dp数组初始化:  dp[0] = 1，理论上也很好解释，装满容量为0的背包，有1种方法，就是装0件物品。
      *          4)dp数组的遍历顺序:
      *          5)dp的举例推导:当nums=[1,1,1,1,1],target=3时
      *                      [0 1 2 3 4]