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* Modify method name to PascalCase(array and linked list) * Modify method name to PascalCase(backtracking) * Modify method name to PascalCase(computational complexity) * Modify method name to PascalCase(divide and conquer) * Modify method name to PascalCase(dynamic programming) * Modify method name to PascalCase(graph) * Modify method name to PascalCase(greedy) * Modify method name to PascalCase(hashing) * Modify method name to PascalCase(heap) * Modify method name to PascalCase(searching) * Modify method name to PascalCase(sorting) * Modify method name to PascalCase(stack and queue) * Modify method name to PascalCase(tree) * local check
71 lines
2.2 KiB
C#
71 lines
2.2 KiB
C#
/**
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* File: coin_change.cs
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* Created Time: 2023-07-12
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* Author: hpstory (hpstory1024@163.com)
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*/
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namespace hello_algo.chapter_dynamic_programming;
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public class coin_change {
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/* 零钱兑换:动态规划 */
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public int CoinChangeDP(int[] coins, int amt) {
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int n = coins.Length;
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int MAX = amt + 1;
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// 初始化 dp 表
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int[,] dp = new int[n + 1, amt + 1];
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// 状态转移:首行首列
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for (int a = 1; a <= amt; a++) {
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dp[0, a] = MAX;
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}
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// 状态转移:其余行列
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for (int i = 1; i <= n; i++) {
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for (int a = 1; a <= amt; a++) {
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if (coins[i - 1] > a) {
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// 若超过背包容量,则不选硬币 i
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dp[i, a] = dp[i - 1, a];
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} else {
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// 不选和选硬币 i 这两种方案的较小值
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dp[i, a] = Math.Min(dp[i - 1, a], dp[i, a - coins[i - 1]] + 1);
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}
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}
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}
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return dp[n, amt] != MAX ? dp[n, amt] : -1;
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}
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/* 零钱兑换:空间优化后的动态规划 */
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public int CoinChangeDPComp(int[] coins, int amt) {
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int n = coins.Length;
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int MAX = amt + 1;
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// 初始化 dp 表
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int[] dp = new int[amt + 1];
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Array.Fill(dp, MAX);
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dp[0] = 0;
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// 状态转移
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for (int i = 1; i <= n; i++) {
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for (int a = 1; a <= amt; a++) {
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if (coins[i - 1] > a) {
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// 若超过背包容量,则不选硬币 i
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dp[a] = dp[a];
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} else {
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// 不选和选硬币 i 这两种方案的较小值
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dp[a] = Math.Min(dp[a], dp[a - coins[i - 1]] + 1);
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}
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}
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}
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return dp[amt] != MAX ? dp[amt] : -1;
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}
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[Test]
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public void Test() {
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int[] coins = { 1, 2, 5 };
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int amt = 4;
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// 动态规划
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int res = CoinChangeDP(coins, amt);
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Console.WriteLine("凑到目标金额所需的最少硬币数量为 " + res);
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// 空间优化后的动态规划
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res = CoinChangeDPComp(coins, amt);
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Console.WriteLine("凑到目标金额所需的最少硬币数量为 " + res);
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}
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}
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