diff --git a/src/main/java/leetcode/editor/cn/CoinChange2.java b/src/main/java/leetcode/editor/cn/CoinChange2.java
new file mode 100644
index 0000000..274438e
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/CoinChange2.java
@@ -0,0 +1,89 @@
+//给定不同面额的硬币和一个总金额。写出函数来计算可以凑成总金额的硬币组合数。假设每一种面额的硬币有无限个。
+//
+//
+//
+//
+//
+//
+// 示例 1:
+//
+// 输入: amount = 5, coins = [1, 2, 5]
+//输出: 4
+//解释: 有四种方式可以凑成总金额:
+//5=5
+//5=2+2+1
+//5=2+1+1+1
+//5=1+1+1+1+1
+//
+//
+// 示例 2:
+//
+// 输入: amount = 3, coins = [2]
+//输出: 0
+//解释: 只用面额2的硬币不能凑成总金额3。
+//
+//
+// 示例 3:
+//
+// 输入: amount = 10, coins = [10]
+//输出: 1
+//
+//
+//
+//
+// 注意:
+//
+// 你可以假设：
+//
+//
+// 0 <= amount (总金额) <= 5000
+// 1 <= coin (硬币面额) <= 5000
+// 硬币种类不超过 500 种
+// 结果符合 32 位符号整数
+//
+// 👍 456 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.Arrays;
+
+//518:零钱兑换 II
+public class CoinChange2 {
+ public static void main(String[] args) {
+ //测试代码
+ Solution solution = new CoinChange2().new Solution();
+ System.out.println(solution.change(5,new int[]{1,2,5}));
+ System.out.println(solution.change(3,new int[]{2}));
+ System.out.println(solution.change(10,new int[]{10}));
+ }
+
+ //力扣代码
+ //leetcode submit region begin(Prohibit modification and deletion)
+ class Solution {
+ public int change(int amount, int[] coins) {
+ int[][] map = new int[coins.length][amount + 1];
+ for (int[] ints : map) {
+ Arrays.fill(ints, -1);
+ }
+ return dfs(coins, 0, amount, map);
+ }
+
+ int dfs(int[] coins, int i, int amount, int[][] map) {
+ if (amount == 0) {
+ return 1;
+ }
+ if (amount < 0 || i == coins.length) {
+ return 0;
+ }
+ if (map[i][amount] != -1) {
+ return map[i][amount];
+ }
+ int res = 0;
+ res += dfs(coins, i + 1, amount, map);
+ res += dfs(coins, i, amount - coins[i], map);
+ return map[i][amount] = res;
+ }
+ }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/CoinChange2.md b/src/main/java/leetcode/editor/cn/CoinChange2.md
new file mode 100644
index 0000000..45411f4
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/CoinChange2.md
@@ -0,0 +1,44 @@
+给定不同面额的硬币和一个总金额。写出函数来计算可以凑成总金额的硬币组合数。假设每一种面额的硬币有无限个。
+
+
+
+
+
+示例 1:
+
+输入: amount = 5, coins = [1, 2, 5]
+输出: 4
+解释: 有四种方式可以凑成总金额:
+5=5
+5=2+2+1
+5=2+1+1+1
+5=1+1+1+1+1
+
+
+示例 2:
+
+输入: amount = 3, coins = [2]
+输出: 0
+解释: 只用面额2的硬币不能凑成总金额3。
+
+
+示例 3:
+
+输入: amount = 10, coins = [10]
+输出: 1
+
+
+
+
+注意:
+
+你可以假设：
+
+
+ - 0 <= amount (总金额) <= 5000
+ - 1 <= coin (硬币面额) <= 5000
+ - 硬币种类不超过 500 种
+ - 结果符合 32 位符号整数
+
+👍 456👎 0
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