From 1ababd77765b8752c565058c24968ebcc34af819 Mon Sep 17 00:00:00 2001
From: LouisFonda <yigencong@yahoo.com>
Date: Tue, 24 Jun 2025 01:31:57 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=E5=8A=A8=E6=80=81=E8=A7=84?=
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---
 .../dynamic-planning/332零钱兑换.js           | 51 +++++++++++++++++++
 1 file changed, 51 insertions(+)
 create mode 100644 top-interview-leetcode150/dynamic-planning/332零钱兑换.js

diff --git a/top-interview-leetcode150/dynamic-planning/332零钱兑换.js b/top-interview-leetcode150/dynamic-planning/332零钱兑换.js
new file mode 100644
index 0000000..b2d6ac6
--- /dev/null
+++ b/top-interview-leetcode150/dynamic-planning/332零钱兑换.js
@@ -0,0 +1,51 @@
+/**
+ * @param {number[]} coins
+ * @param {number} amount
+ * @return {number}
+ */
+const coinChange = function (coins, amount) {
+
+};
+
+/*
+这个问题可以抽象成一个完全背包问题，物品的价值就是硬币的面值，物品的重量也是硬币的面值，定义dp[i][j]表示0-i类型的硬币
+中，组合成j所需硬币的数量
+*/
+function f1(coins, amount) {
+  const m = coins.length;
+  const n = amount;
+  const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(Infinity));
+
+  // 初始化：金额为0，硬币数为0
+  for (let i = 0; i <= m; i++) {
+    dp[i][0] = 0;
+  }
+
+  for (let i = 1; i <= m; i++) {
+    const coin = coins[i - 1];
+    for (let j = 1; j <= n; j++) {
+      if (j < coin) {
+        dp[i][j] = dp[i - 1][j]; // 当前硬币无法选，用上一行的结果
+      } else {
+        // 可以选当前硬币（完全背包，不是 i-1）
+        dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - coin] + 1);
+      }
+    }
+  }
+
+  return dp[m][n] === Infinity ? -1 : dp[m][n];
+}
+
+// 上面的代码可以使用一维dp来实现，dp[i]表示找i元需要最少的硬币
+function coinChange(coins, amount) {
+  const dp = Array(amount + 1).fill(Infinity);
+  dp[0] = 0; // 金额为 0 时需要 0 个硬币
+
+  for (let coin of coins) {
+    for (let j = coin; j <= amount; j++) {
+      dp[j] = Math.min(dp[j], dp[j - coin] + 1);
+    }
+  }
+
+  return dp[amount] === Infinity ? -1 : dp[amount];
+}
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