feat: add solutions to leetcode problem: No.0322. Coin Change

Author: maolonglong

solution/0300-0399/0322.Coin Change/README.md

@@ -16,7 +16,7 @@
 
 <pre>
 <strong>输入：</strong>coins = <code>[1, 2, 5]</code>, amount = <code>11</code>
-<strong>输出：</strong><code>3</code> 
+<strong>输出：</strong><code>3</code>
 <strong>解释：</strong>11 = 5 + 5 + 1</pre>
 
 <p><strong>示例 2：</strong></p>
@@ -61,28 +61,47 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+类似完全背包的思路，硬币数量不限，求凑成总金额所需的最少的硬币个数。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        dp = [amount + 1 for i in range(amount + 1)]
+        dp[0] = 0
+        for coin in coins:
+            for j in range(coin, amount + 1):
+                dp[j] = min(dp[j], dp[j - coin] + 1)
+        return -1 if dp[amount] > amount else dp[amount]
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int[] dp = new int[amount + 1];
+        Arrays.fill(dp, amount + 1);
+        dp[0] = 0;
+        for (int coin : coins) {
+            for (int j = coin; j <= amount; j++) {
+                dp[j] = Math.min(dp[j], dp[j - coin] + 1);
+            }
+        }
+        return dp[amount] > amount ? -1 : dp[amount];
+    }
+}
 ```
 
 ### **JavaScript**
 
-动态规划法。
-
 ```js
 /**
  * @param {number[]} coins

solution/0300-0399/0322.Coin Change/README_EN.md

@@ -59,24 +59,43 @@
 
 ## Solutions
 
+Similar to the idea of ​​a complete backpack, there is no limit to the number of coins. Find the minimum number of coins required to make up the total amount.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        dp = [amount + 1 for i in range(amount + 1)]
+        dp[0] = 0
+        for coin in coins:
+            for j in range(coin, amount + 1):
+                dp[j] = min(dp[j], dp[j - coin] + 1)
+        return -1 if dp[amount] > amount else dp[amount]
 ```
 
 ### **Java**
 
 ```java
-
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int[] dp = new int[amount + 1];
+        Arrays.fill(dp, amount + 1);
+        dp[0] = 0;
+        for (int coin : coins) {
+            for (int j = coin; j <= amount; j++) {
+                dp[j] = Math.min(dp[j], dp[j - coin] + 1);
+            }
+        }
+        return dp[amount] > amount ? -1 : dp[amount];
+    }
+}
 ```
 
 ### **JavaScript**
 
-Dynamic programming.
-
 ```js
 /**
  * @param {number[]} coins

solution/0300-0399/0322.Coin Change/Solution.java

@@ -1,16 +1,13 @@
 class Solution {
-	public int coinChange(int[] coins, int amount) {
-		int n = coins.length;
-		int[] dp = new int[amount + 1];
-		Arrays.fill(dp, amount + 1);
-		dp[0] = 0;
-		for (int i = 1; i <= amount; i++) {
-			for (int j = 0; j < n; j++) {
-				if (i >= coins[j]) {
-					dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
-				}
-			}
-		}
-		return dp[amount] > amount ? -1 : dp[amount];
-	}
-}
+    public int coinChange(int[] coins, int amount) {
+        int[] dp = new int[amount + 1];
+        Arrays.fill(dp, amount + 1);
+        dp[0] = 0;
+        for (int coin : coins) {
+            for (int j = coin; j <= amount; j++) {
+                dp[j] = Math.min(dp[j], dp[j - coin] + 1);
+            }
+        }
+        return dp[amount] > amount ? -1 : dp[amount];
+    }
+}

solution/0300-0399/0322.Coin Change/Solution.py

@@ -0,0 +1,8 @@
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        dp = [amount + 1 for i in range(amount + 1)]
+        dp[0] = 0
+        for coin in coins:
+            for j in range(coin, amount + 1):
+                dp[j] = min(dp[j], dp[j - coin] + 1)
+        return -1 if dp[amount] > amount else dp[amount]