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Commit efd1993

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feat: add solutions to leetcode problem: No.0322. Coin Change
1 parent 09a0eb6 commit efd1993

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4 files changed

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-24
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‎solution/0300-0399/0322.Coin Change/README.md

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<pre>
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<strong>输入:</strong>coins = <code>[1, 2, 5]</code>, amount = <code>11</code>
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<strong>输出:</strong><code>3</code>
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<strong>输出:</strong><code>3</code>
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<strong>解释:</strong>11 = 5 + 5 + 1</pre>
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<p><strong>示例 2:</strong></p>
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<!-- 这里可写通用的实现逻辑 -->
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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。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def coinChange(self, coins: List[int], amount: int) -> int:
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dp = [amount + 1 for i in range(amount + 1)]
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dp[0] = 0
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for coin in coins:
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for j in range(coin, amount + 1):
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dp[j] = min(dp[j], dp[j - coin] + 1)
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return -1 if dp[amount] > amount else dp[amount]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int coinChange(int[] coins, int amount) {
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int[] dp = new int[amount + 1];
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Arrays.fill(dp, amount + 1);
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dp[0] = 0;
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for (int coin : coins) {
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for (int j = coin; j <= amount; j++) {
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dp[j] = Math.min(dp[j], dp[j - coin] + 1);
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}
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}
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return dp[amount] > amount ? -1 : dp[amount];
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}
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}
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```
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### **JavaScript**
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动态规划法。
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```js
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/**
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* @param {number[]} coins

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

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## Solutions
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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.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def coinChange(self, coins: List[int], amount: int) -> int:
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dp = [amount + 1 for i in range(amount + 1)]
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dp[0] = 0
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for coin in coins:
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for j in range(coin, amount + 1):
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dp[j] = min(dp[j], dp[j - coin] + 1)
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return -1 if dp[amount] > amount else dp[amount]
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```
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### **Java**
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```java
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class Solution {
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public int coinChange(int[] coins, int amount) {
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int[] dp = new int[amount + 1];
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Arrays.fill(dp, amount + 1);
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dp[0] = 0;
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for (int coin : coins) {
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for (int j = coin; j <= amount; j++) {
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dp[j] = Math.min(dp[j], dp[j - coin] + 1);
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}
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}
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return dp[amount] > amount ? -1 : dp[amount];
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}
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}
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```
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### **JavaScript**
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Dynamic programming.
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```js
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/**
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* @param {number[]} coins
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class Solution {
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public int coinChange(int[] coins, int amount) {
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int n = coins.length;
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int[] dp = new int[amount + 1];
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Arrays.fill(dp, amount + 1);
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dp[0] = 0;
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for (int i = 1; i <= amount; i++) {
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for (int j = 0; j < n; j++) {
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if (i >= coins[j]) {
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dp[i] = Math.min(dp[i], dp[i - coins[j]] + 1);
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}
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}
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}
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return dp[amount] > amount ? -1 : dp[amount];
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}
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}
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public int coinChange(int[] coins, int amount) {
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int[] dp = new int[amount + 1];
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Arrays.fill(dp, amount + 1);
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dp[0] = 0;
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for (int coin : coins) {
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for (int j = coin; j <= amount; j++) {
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dp[j] = Math.min(dp[j], dp[j - coin] + 1);
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}
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}
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return dp[amount] > amount ? -1 : dp[amount];
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}
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}
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class Solution:
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def coinChange(self, coins: List[int], amount: int) -> int:
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dp = [amount + 1 for i in range(amount + 1)]
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dp[0] = 0
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for coin in coins:
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for j in range(coin, amount + 1):
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dp[j] = min(dp[j], dp[j - coin] + 1)
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return -1 if dp[amount] > amount else dp[amount]

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