From af4056827594bdc1db564a91cc86d196e5a48a6d Mon Sep 17 00:00:00 2001
From: EmmIons <48384706+EmmIons@users.noreply.github.com>
Date: Sat, 20 Jan 2024 18:02:03 +0800
Subject: [PATCH] =?UTF-8?q?Update=200518.=E9=9B=B6=E9=92=B1=E5=85=91?=
 =?UTF-8?q?=E6=8D=A2II.md?=
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java版本二维dp数组代码逻辑优化, 原版本的二维dp数组和一维dp数组的逻辑并不对应.
---
 problems/0518.零钱兑换II.md | 28 +++++++++++++++++-----------
 1 file changed, 17 insertions(+), 11 deletions(-)

diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md
index 5c6efdcb..fa104541 100644
--- a/problems/0518.零钱兑换II.md
+++ b/problems/0518.零钱兑换II.md
@@ -224,20 +224,26 @@ class Solution {
 // 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;
+        int[][] dp = new int[coins.length][amount+1];
+
+        // 初始化边界值
+        for(int i = 0; i < coins.length; i++){
+            // 第一列的初始值为1
+            dp[i][0] = 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]];
-                }
+        for(int j = coins[0]; j <= amount; j++){
+            // 初始化第一行
+            dp[0][j] += dp[0][j-coins[0]];
+        }
+        
+        for(int i = 1; i < coins.length; i++){
+            for(int j = 1; j <= amount; j++){
+                if(j < coins[i]) dp[i][j] = dp[i-1][j];
+                else dp[i][j] = dp[i][j-coins[i]] + dp[i-1][j];
             }
         }
-        return dp[coins.length - 1][amount];
+
+        return dp[coins.length-1][amount];
     }
 }
 ```