89 lines
2.1 KiB
Java
89 lines
2.1 KiB
Java
//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。
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//
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//
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//
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//
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//
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//
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// 示例 1:
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//
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// 输入: amount = 5, coins = [1, 2, 5]
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//输出: 4
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//解释: 有四种方式可以凑成总金额:
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//5=5
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//5=2+2+1
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//5=2+1+1+1
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//5=1+1+1+1+1
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//
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//
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// 示例 2:
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//
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// 输入: amount = 3, coins = [2]
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//输出: 0
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//解释: 只用面额2的硬币不能凑成总金额3。
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//
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//
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// 示例 3:
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//
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// 输入: amount = 10, coins = [10]
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//输出: 1
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//
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//
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//
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//
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// 注意:
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//
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// 你可以假设：
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//
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//
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// 0 <= amount (总金额) <= 5000
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// 1 <= coin (硬币面额) <= 5000
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// 硬币种类不超过 500 种
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// 结果符合 32 位符号整数
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//
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// 👍 456 👎 0
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package leetcode.editor.cn;
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import java.util.Arrays;
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//518:零钱兑换 II
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public class CoinChange2 {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new CoinChange2().new Solution();
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System.out.println(solution.change(5,new int[]{1,2,5}));
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System.out.println(solution.change(3,new int[]{2}));
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System.out.println(solution.change(10,new int[]{10}));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int change(int amount, int[] coins) {
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int[][] map = new int[coins.length][amount + 1];
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for (int[] ints : map) {
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Arrays.fill(ints, -1);
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}
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return dfs(coins, 0, amount, map);
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}
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int dfs(int[] coins, int i, int amount, int[][] map) {
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if (amount == 0) {
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return 1;
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}
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if (amount < 0 || i == coins.length) {
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return 0;
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}
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if (map[i][amount] != -1) {
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return map[i][amount];
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}
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int res = 0;
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res += dfs(coins, i + 1, amount, map);
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res += dfs(coins, i, amount - coins[i], map);
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return map[i][amount] = res;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |