From 334f4a27de2ecc00134c6cf287f658758f856e6f Mon Sep 17 00:00:00 2001
From: hoxiansen <1365171821@qq.com>
Date: Tue, 28 Mar 2023 09:40:19 +0000
Subject: [PATCH] =?UTF-8?q?Update=200518.=E9=9B=B6=E9=92=B1=E5=85=91?=
 =?UTF-8?q?=E6=8D=A2II.md=20=E8=A1=A5=E5=85=85=E4=BA=8C=E7=BB=B4dp?=
 =?UTF-8?q?=E6=95=B0=E7=BB=84=E7=9A=84Java=E4=BB=A3=E7=A0=81?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

---
 problems/0518.零钱兑换II.md | 21 +++++++++++++++++++++
 1 file changed, 21 insertions(+)

diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md
index 3fd03e16..eb5a844c 100644
--- a/problems/0518.零钱兑换II.md
+++ b/problems/0518.零钱兑换II.md
@@ -215,6 +215,27 @@ class Solution {
     }
 }
 ```
+```Java
+// 二维dp数组版本，方便理解
+class Solution {
+    public int change(int amount, int[] coins) {
+        int[][] dp = new int[coins.length][amount + 1];
+        // 只有一种硬币的情况
+        for (int i = 0; i <= amount; i += coins[0]) {
+            dp[0][i] = 1;
+        }
+        for (int i = 1; i < coins.length; i++) {
+            for (int j = 0; j <= amount; j++) {
+                // 第i种硬币使用0~k次，求和
+                for (int k = 0; k * coins[i] <= j; k++) {
+                    dp[i][j] += dp[i - 1][j - k * coins[i]];
+                }
+            }
+        }
+        return dp[coins.length - 1][amount];
+    }
+}
+```
 
 Python：