fix: update leetcode solution: 0322.Coin Change

更新LeetCode题解

Author: yanglbme

README.md

@@ -25,8 +25,8 @@
 
 ## 题解
 - [LeetCode](/solution/README.md)
-- [剑指 Offer（第 2 版）](/lcof/README.md)
-- [程序员面试金典（第 6 版）](/lcci/README.md)
+- [LeetCode 《剑指 Offer（第 2 版）》](/lcof/README.md)
+- [LeetCode 《程序员面试金典（第 6 版）》](/lcci/README.md)
 
 ## 维护者
 [Yang Libin](https://github.com/yanglbme): GitHub 技术社区 [@Doocs](https://github.com/doocs) 创建者；[@TheAlgorithms](https://github.com/TheAlgorithms) 组织成员。

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

@@ -39,19 +39,26 @@
 ```
 
 ### JavaScript
-* Dynamic programming:
-```JavaScript
-var coinChange = function(coins, amount) {
-  var dp = Array(amount + 1).fill(amount + 1);
-  dp[0] = 0;
-  for (var i = 1; i <= amount; i++) {
-      for (var j = 0; j < coins.length; j++) {
-          if (coins[j] <= i) {
-              dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
-          }
-      }
-  }
-  
-  return dp[amount] > amount ? -1 : dp[amount];
+动态规划法。
+
+```javascript
+/**
+ * @param {number[]} coins
+ * @param {number} amount
+ * @return {number}
+ */
+var coinChange = function (coins, amount) {
+    var dp = Array(amount + 1).fill(amount + 1);
+    dp[0] = 0;
+    for (var i = 1; i <= amount; i++) {
+        for (var j = 0; j < coins.length; j++) {
+            if (coins[j] <= i) {
+                dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
+            }
+        }
+    }
+
+    return dp[amount] > amount ? -1 : dp[amount];
 };
+
 ```

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

@@ -1,43 +1,79 @@
-# [322. Coin Change](https://leetcode.com/problems/coin-change)
-
-## Description
-<p>You are given coins of different denominations and a total amount of money <i>amount</i>. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return <code>-1</code>.</p>
-
-<p><b>Example 1:</b></p>
-
-<pre>
-<strong>Input: </strong>coins = <code>[1, 2, 5]</code>, amount = <code>11</code>
-<strong>Output: </strong><code>3</code> 
-<strong>Explanation:</strong> 11 = 5 + 5 + 1</pre>
-
-<p><b>Example 2:</b></p>
-
-<pre>
-<strong>Input: </strong>coins = <code>[2]</code>, amount = <code>3</code>
-<strong>Output: </strong>-1
-</pre>
-
-<p><b>Note</b>:<br />
-You may assume that you have an infinite number of each kind of coin.</p>
-
-
-
-## Solutions
-
-
-### Python3
-
-```python
-
-```
-
-### Java
-
-```java
-
-```
-
-### ...
-```
-
-```
+# [322. Coin Change](https://leetcode.com/problems/coin-change)
+
+## Description
+<p>You are given coins of different denominations and a total amount of money <i>amount</i>. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return <code>-1</code>.</p>
+
+
+
+<p><b>Example 1:</b></p>
+
+
+
+<pre>
+
+<strong>Input: </strong>coins = <code>[1, 2, 5]</code>, amount = <code>11</code>
+
+<strong>Output: </strong><code>3</code> 
+
+<strong>Explanation:</strong> 11 = 5 + 5 + 1</pre>
+
+
+
+<p><b>Example 2:</b></p>
+
+
+
+<pre>
+
+<strong>Input: </strong>coins = <code>[2]</code>, amount = <code>3</code>
+
+<strong>Output: </strong>-1
+
+</pre>
+
+
+
+<p><b>Note</b>:<br />
+
+You may assume that you have an infinite number of each kind of coin.</p>
+
+
+
+
+## Solutions
+
+
+### Python3
+
+```python
+
+```
+
+### Java
+
+```java
+
+```
+
+### JavaScript
+Dynamic programming.
+
+```javascript
+/**
+ * @param {number[]} coins
+ * @param {number} amount
+ * @return {number}
+ */
+var coinChange = function (coins, amount) {
+    var dp = Array(amount + 1).fill(amount + 1);
+    dp[0] = 0;
+    for (var i = 1; i <= amount; i++) {
+        for (var j = 0; j < coins.length; j++) {
+            if (coins[j] <= i) {
+                dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
+            }
+        }
+    }
+
+    return dp[amount] > amount ? -1 : dp[amount];
+};

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

@@ -3,16 +3,16 @@
  * @param {number} amount
  * @return {number}
  */
-var coinChange = function(coins, amount) {
-  var dp = Array(amount + 1).fill(amount + 1);
-  dp[0] = 0;
-  for (var i = 1; i <= amount; i++) {
-      for (var j = 0; j < coins.length; j++) {
-          if (coins[j] <= i) {
-              dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
-          }
-      }
-  }
-  
-  return dp[amount] > amount ? -1 : dp[amount];
+var coinChange = function (coins, amount) {
+    var dp = Array(amount + 1).fill(amount + 1);
+    dp[0] = 0;
+    for (var i = 1; i <= amount; i++) {
+        for (var j = 0; j < coins.length; j++) {
+            if (coins[j] <= i) {
+                dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
+            }
+        }
+    }
+
+    return dp[amount] > amount ? -1 : dp[amount];
 };