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518:零钱兑换 II

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huangge1199 2021-06-10 14:22:10 +08:00
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src/main/java/leetcode/editor/cn/CoinChange2.java Normal file
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//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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src/main/java/leetcode/editor/cn/CoinChange2.md Normal file
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<p>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。&nbsp;</p>
<p>&nbsp;</p>
<ul>
</ul>
<p><strong>示例 1:</strong></p>
<pre><strong>输入:</strong> amount = 5, coins = [1, 2, 5]
<strong>输出:</strong> 4
<strong>解释:</strong> 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
</pre>
<p><strong>示例 2:</strong></p>
<pre><strong>输入:</strong> amount = 3, coins = [2]
<strong>输出:</strong> 0
<strong>解释:</strong> 只用面额2的硬币不能凑成总金额3。
</pre>
<p><strong>示例 3:</strong></p>
<pre><strong>输入:</strong> amount = 10, coins = [10]
<strong>输出:</strong> 1
</pre>
<p>&nbsp;</p>
<p><strong>注意</strong><strong>:</strong></p>
<p>你可以假设：</p>
<ul>
<li>0 &lt;= amount (总金额) &lt;= 5000</li>
<li>1 &lt;= coin (硬币面额)&nbsp;&lt;= 5000</li>
<li>硬币种类不超过 500 种</li>
<li>结果符合 32 位符号整数</li>
</ul>
<div><li>👍 456</li><li>👎 0</li></div>